Diophantine equations

## Sequences, series and numbers generated by diophantine equations and their applications by Jamel Ghanouchi

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## Abstract

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Our purpose in this article which has been published is to show how much diophantine equations are rich in analytic applications. Effectively, those equations allow to build amazing sequences, series and numbers. The question of the proof of some theorems remains of course, we will see it in this communication. We will make also an allusion to the very known Fermat numbers ( 2 2 n {\displaystyle 2^{2^{n}}} ). We will see how this problem of the proof is actual and how it can be solved using amazing sequences and series.

## The Ghanouchi's sequences

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Let us begin by Fermat equation (E), it is

U n = X n + Y n {\displaystyle U^{n}=X^{n}+Y^{n}}

with GCD(X,Y)=1

We pose

u = U 2 n {\displaystyle u=U^{2n}}

x = U n X n {\displaystyle x=U^{n}X^{n}}

y = U n Y n {\displaystyle y=U^{n}Y^{n}}

z = X n Y n {\displaystyle z=X^{n}Y^{n}}

then

u = U 2 n = U n ( X n + Y n ) = x + y ( 1 ) {\displaystyle u=U^{2n}=U^{n}(X^{n}+Y^{n})=x+y\quad \quad {(1)}}

and

1 z = 1 X n Y n = U 2 n U n X n U n Y n = u x y = x + y x y = 1 x + 1 y ( 2 ) {\displaystyle {\frac {1}{z}}={\frac {1}{X^{n}Y^{n}}}={\frac {U^{2n}}{U^{n}X^{n}U^{n}Y^{n}}}={\frac {u}{xy}}={\frac {x+y}{xy}}={\frac {1}{x}}+{\frac {1}{y}}\quad \quad {(2)}}

If U, X, Y are integers verifying equation (E), then u, x, y, z as defined verify

## LEMMA 1

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u = x + y ( 1 ) {\displaystyle u=x+y\quad \quad {(1)}}

1 z = 1 x + 1 y ( 2 ) {\displaystyle {\frac {1}{z}}={\frac {1}{x}}+{\frac {1}{y}}\quad \quad {(2)}}

Firstly

z = X n Y n {\displaystyle z=X^{n}Y^{n}}

u = ( X n + Y n ) 2 {\displaystyle u=(X^{n}+Y^{n})^{2}}

x = X n ( X n + Y n ) {\displaystyle x=X^{n}(X^{n}+Y^{n})}

y = Y n ( X n + Y n ) {\displaystyle y=Y^{n}(X^{n}+Y^{n})}

We pose

x 1 = x {\displaystyle x_{1}=x}

and

y 1 = y {\displaystyle y_{1}=y}

but

∀ x 1 , y 1 {\displaystyle \forall {x_{1},y_{1}}}

∃ z 1 {\displaystyle \exists {z_{1}}}

verifying

1 z 1 = 1 x 1 + 1 y 1 {\displaystyle {\frac {1}{z_{1}}}={\frac {1}{x_{1}}}+{\frac {1}{y_{1}}}}

and

z 1 = x y x + y = z {\displaystyle z_{1}={\frac {xy}{x+y}}=z}

then

( x 1 + y 1 ) z 1 = x 1 y 1 {\displaystyle (x_{1}+y_{1})z_{1}=x_{1}y_{1}}

x 1 ( y 1 − z 1 ) = z 1 y 1 {\displaystyle x_{1}(y_{1}-z_{1})=z_{1}y_{1}}

we pose

y 2 = y 1 − z 1 = z 1 y 1 x 1 {\displaystyle y_{2}=y_{1}-z_{1}={\frac {z_{1}y_{1}}{x_{1}}}}

and

y 1 ( x 1 − z 1 ) = z 1 x 1 {\displaystyle y_{1}(x_{1}-z_{1})=z_{1}x_{1}}

also

x 2 = x 1 − z 1 = z 1 x 1 y 1 {\displaystyle x_{2}=x_{1}-z_{1}={\frac {z_{1}x_{1}}{y_{1}}}}

and

x 2 y 2 = z 1 2 {\displaystyle x_{2}y_{2}=z_{1}^{2}}

which means

x 1 = x 2 + z 1 = x 2 + x 2 y 2 {\displaystyle x_{1}=x_{2}+z_{1}=x_{2}+{\sqrt {x_{2}y_{2}}}}

and

y 1 = y 2 + z 1 = y 2 + x 2 y 2 {\displaystyle y_{1}=y_{2}+z_{1}=y_{2}+{\sqrt {x_{2}y_{2}}}}

and

u 1 = u = ( x 1 + y 1 ) = ( x 2 + y 2 ) 2 > x 2 + y 2 > 0 {\displaystyle u_{1}=u=(x_{1}+y_{1})=({\sqrt {x_{2}}}+{\sqrt {y_{2}}})^{2}>x_{2}+y_{2}>0}

( x 1 + y 1 ) {\displaystyle (x_{1}+y_{1})} is an integer

x 1 = x 2 ( x 2 + y 2 ) > x 2 > 0 {\displaystyle x_{1}={\sqrt {x_{2}}}({\sqrt {x_{2}}}+{\sqrt {y_{2}}})>x_{2}>0}

x 1 {\displaystyle x_{1}} is an integer

y 1 = y 2 ( x 2 + y 2 ) > y 2 > 0 {\displaystyle y_{1}={\sqrt {y_{2}}}({\sqrt {x_{2}}}+{\sqrt {y_{2}}})>y_{2}>0}

y 1 {\displaystyle y_{1}} is an integer

z 1 = x 1 y 1 x 1 + y 1 = X a Y b = x 2 y 2 > z 2 = x 2 y 2 x 2 + y 2 > 0 {\displaystyle z_{1}={\frac {x_{1}y_{1}}{x_{1}+y_{1}}}=X^{a}Y^{b}={\sqrt {x_{2}y_{2}}}>z_{2}={\frac {x_{2}y_{2}}{x_{2}+y_{2}}}>0}

z 2 {\displaystyle z_{2}} is rational, because

∀ x 2 , y 2 {\displaystyle \forall {x_{2},y_{2}}} rationals

∃ z 2 {\displaystyle \exists {z_{2}}} rational verifying

1 z 2 = 1 x 2 + 1 y 2 {\displaystyle {\frac {1}{z_{2}}}={\frac {1}{x_{2}}}+{\frac {1}{y_{2}}}}

until infinity. For i

( x i + y i ) = ( x i + 1 + y i + 1 ) 2 > x i + 1 + y i + 1 > 0 {\displaystyle (x_{i}+y_{i})=({\sqrt {x_{i+1}}}+{\sqrt {y_{i+1}}})^{2}>x_{i+1}+y_{i+1}>0}

x i + y i {\displaystyle x_{i}+y_{i}} is rational for i>1

x i = x i + 1 ( x i + 1 + y i + 1 ) > x i + 1 > 0 {\displaystyle x_{i}={\sqrt {x_{i+1}}}({\sqrt {x_{i+1}}}+{\sqrt {y_{i+1}}})>x_{i+1}>0}

x i {\displaystyle x_{i}} is rational for i>1

y i = y i + 1 ( x i + 1 + y i + 1 ) > y i + 1 > 0 {\displaystyle y_{i}={\sqrt {y_{i+1}}}({\sqrt {x_{i+1}}}+{\sqrt {y_{i+1}}})>y_{i+1}>0}

y i {\displaystyle y_{i}} is rational for i>1

z i = x i y i x i + y i = x i + 1 y i + 1 > z i + 1 = x i + 1 y i + 1 x i + 1 + y i + 1 > 0 {\displaystyle z_{i}={\frac {x_{i}y_{i}}{x_{i}+y_{i}}}={\sqrt {x_{i+1}y_{i+1}}}>z_{i+1}={\frac {x_{i+1}y_{i+1}}{x_{i+1}+y_{i+1}}}>0}

z i {\displaystyle z_{i}} is rational for i>1, and

1 z i + 1 = 1 x i + 1 + 1 y i + 1 {\displaystyle {\frac {1}{z_{i+1}}}={\frac {1}{x_{i+1}}}+{\frac {1}{y_{i+1}}}}

( x i + y i ) = ( x i + 1 + y i + 1 ) 2 {\displaystyle (x_{i}+y_{i})=({\sqrt {x_{i+1}}}+{\sqrt {y_{i+1}}})^{2}}

## LEMMA 2

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x i {\displaystyle x_{i}} and  y i {\displaystyle y_{i}} have expressions

x i = x 2 i − 1 ∏ j = 0 j = i − 2 ( x 2 j + y 2 j ) − 1 ( H ) {\displaystyle x_{i}=x^{2^{i-1}}\prod _{j=0}^{j={i-2}}{(x^{2^{j}}+y^{2^{j}})^{-1}}\quad \quad {(H)}}

y i = y 2 i − 1 ∏ j = 0 j = i − 2 ( x 2 j + y 2 j ) − 1 ( H ′ ) {\displaystyle y_{i}=y^{2^{i-1}}\prod _{j=0}^{j={i-2}}{(x^{2^{j}}+y^{2^{j}})^{-1}}\quad \quad {(H')}}

## Proof of lemma 2

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By induction

x = x 2 ( x 2 + y 2 ) = x 2 ( x + y ) 1 2 {\displaystyle x={\sqrt {x_{2}}}({\sqrt {x_{2}}}+{\sqrt {y_{2}}})={\sqrt {x_{2}}}(x+y)^{\frac {1}{2}}}

x 2 = x 2 x + y {\displaystyle x_{2}={\frac {x^{2}}{x+y}}}

also

y 2 = y 2 x + y {\displaystyle y_{2}={\frac {y^{2}}{x+y}}}

it is verified for i=2. We suppose (H) and (H') true for i, then

x i = x i + 1 ( x i + 1 + y i + 1 ) = x i + 1 ( x i + y i ) 1 2 {\displaystyle x_{i}={\sqrt {x_{i+1}}}({\sqrt {x_{i+1}}}+{\sqrt {y_{i+1}}})={\sqrt {x_{i+1}}}(x_{i}+y_{i})^{\frac {1}{2}}}

and

x i + 1 = x i 2 ( x i + y i ) − 1 {\displaystyle x_{i+1}=x_{i}^{2}{(x_{i}+y_{i})}^{-1}}

but (H) and (H'), then

x i + 1 = x 2 i ∏ j = 0 j = i − 2 ( x 2 j + y 2 j ) − 2 ( x 2 i − 1 + y 2 i − 1 ) − 1 ∏ j = 0 j = i − 2 ( x 2 j + y 2 j ) {\displaystyle x_{i+1}=x^{2^{i}}\prod _{j=0}^{j=i-2}{(x^{2^{j}}+y^{2^{j}})^{-2}}(x^{2^{i-1}}+y^{2^{i-1}})^{-1}\prod _{j=0}^{j={i-2}}{(x^{2^{j}}+y^{2^{j}})}}

= x 2 i ∏ j = 0 j = i − 1 ( x 2 j + y 2 j ) − 1 {\displaystyle =x^{2^{i}}\prod _{j=0}^{j={i-1}}{(x^{2^{j}}+y^{2^{j}})^{-1}}}

and it is true for i+1, also for  y i {\displaystyle y_{i}}

x i {\displaystyle x_{i}} and  y i {\displaystyle y_{i}} can be written as it follows

x i = x 2 i − 1 ∏ j = 0 j = i − 2 ( x 2 j + y 2 j ) − 1 {\displaystyle x_{i}=x^{2^{i-1}}\prod _{j=0}^{j=i-2}{(x^{2^{j}}+y^{2^{j}})^{-1}}}

y i = y 2 i − 1 ∏ j = 0 j = i − 2 ( x 2 j + y 2 j ) − 1 {\displaystyle y_{i}=y^{2^{i-1}}\prod _{j=0}^{j=i-2}{(x^{2^{j}}+y^{2^{j}})^{-1}}}

∀ i > 1 {\displaystyle \forall {i>1}}

but,  ∀ a , b {\displaystyle \forall {a,b}}

a − b = ( a 2 i − 1 − b 2 i − 1 ) ∏ j = 0 j = i − 2 ( a 2 j + b 2 j ) − 1 ( E ) {\displaystyle a-b=(a^{2^{i-1}}-b^{2^{i-1}})\prod _{j=0}^{j={i-2}}{(a^{2^{j}}+b^{2^{j}})^{-1}}\quad \quad {(E)}}

x − y = ( x 2 i − 1 − y 2 i − 1 ) ∏ j = 0 j = i − 2 ( x 2 j + y 2 j ) − 1 ( E ′ ) {\displaystyle x-y=(x^{2^{i-1}}-y^{2^{i-1}})\prod _{j=0}^{j={i-2}}{(x^{2^{j}}+y^{2^{j}})^{-1}}\quad \quad {(E')}}

∏ j = 0 j = i − 2 ( x 2 j + y 2 j ) = ( x 2 i − 1 − y 2 i − 1 ) x − y {\displaystyle \prod _{j=0}^{j={i-2}}{(x^{2^{j}}+y^{2^{j}})}={\frac {(x^{2^{i-1}}-y^{2^{i-1}})}{x-y}}}

the expressions of the sequences become, for

x ≠ y {\displaystyle x\neq {y}}

x i ≠ y i {\displaystyle x_{i}\neq {y_{i}}}

## LEMMA 3

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x i = x 2 i − 1 x 2 i − 1 − y 2 i − 1 ( x − y ) {\displaystyle x_{i}={\frac {x^{2^{i-1}}}{x^{2^{i-1}}-y^{2^{i-1}}}}(x-y)}

= U n X n 2 i − 1 X n 2 i − 1 − Y n 2 i − 1 ( X n − Y n ) {\displaystyle =U^{n}{\frac {X^{n{2^{i-1}}}}{X^{n{2^{i-1}}}-Y^{n{2^{i-1}}}}}(X^{n}-Y^{n})}

and

y i = y 2 i − 1 x 2 i − 1 − y 2 i − 1 ( x − y ) {\displaystyle y_{i}={\frac {y^{2^{i-1}}}{x^{2^{i-1}}-y^{2^{i-1}}}}(x-y)}

= U n Y n 2 i − 1 X n 2 i − 1 − Y n 2 i − 1 ( X n − Y n ) {\displaystyle =U^{n}{\frac {Y^{n{2^{i-1}}}}{X^{n{2^{i-1}}}-Y^{n{2^{i-1}}}}}(X^{n}-Y^{n})}

u i = x i + y i = U n X n 2 i − 1 + Y n 2 i − 1 X n 2 i − 1 − Y n 2 i − 1 ( X n − Y n ) {\displaystyle u_{i}=x_{i}+y_{i}=U^{n}{\frac {X^{n{2^{i-1}}}+Y^{n{2^{i-1}}}}{X^{n{2^{i-1}}}-Y^{n{2^{i-1}}}}}(X^{n}-Y^{n})}

## LEMMA 4

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The equations (1) have (2) a constant

x i − y i = x − y {\displaystyle x_{i}-y_{i}=x-y}

## THE GHANOUCHI'S THEOREM

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The only solution of equations

u = x + y , ( ( 1 ) ) {\displaystyle u=x+y,\quad ((1))}

and

1 z = 1 x + 1 y , ( 2 ) {\displaystyle {\frac {1}{z}}={\frac {1}{x}}+{\frac {1}{y}},\quad {(2)}}

x y ( x − y ) = 0 {\displaystyle xy(x-y)=0}

## Proof of Ghanouchi's theorem

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x i = x 2 i − 1 x 2 i − 1 − y 2 i − 1 ( x − y ) {\displaystyle x_{i}={\frac {x^{2^{i-1}}}{x^{2^{i-1}}-y^{2^{i-1}}}}(x-y)}

and

y i = y 2 i − 1 x 2 i − 1 − y 2 i − 1 ( x − y ) {\displaystyle y_{i}={\frac {y^{2^{i-1}}}{x^{2^{i-1}}-y^{2^{i-1}}}}(x-y)}

x > y ⇒ lim i ⟶ ∞ ( x i ) = x − y , lim i ⟶ ∞ ( y i ) = 0 {\displaystyle x>y\Rightarrow {\lim _{i\longrightarrow {\infty }}{(x_{i})}=x-y,\quad \quad {\lim _{i\longrightarrow {\infty }}{(y_{i})}=0}}}

and

y > x ⇒ lim i ⟶ ∞ ( y i ) = y − x , lim i ⟶ ∞ ( x i ) = 0 {\displaystyle y>x\Rightarrow {\lim _{i\longrightarrow {\infty }}{(y_{i})}=y-x,\quad \quad {\lim _{i\longrightarrow {\infty }}{(x_{i})}=0}}}

We will give several proofs that  x y ( x − y ) = 0 {\displaystyle xy(x-y)=0} is the solution with the series. We recapitulate

x i > x i + 1 , y i > y i + 1 ⇒ x = y {\displaystyle x_{i}>x_{i+1},\quad {y_{i}>y_{i+1}}\Rightarrow {x=y}}

x i = x i + 1 = x i + 1 + x i + 1 y i + 1 ⇒ x y = 0 {\displaystyle x_{i}=x_{i+1}=x_{i+1}+{\sqrt {x_{i+1}y_{i+1}}}\Rightarrow {xy=0}}

y i = y i + 1 = y i + 1 + x i + 1 y i + 1 ⇒ x y = 0 {\displaystyle y_{i}=y_{i+1}=y_{i+1}+{\sqrt {x_{i+1}y_{i+1}}}\Rightarrow {xy=0}}

⇒ x y ( x − y ) = 0 {\displaystyle \Rightarrow {xy(x-y)=0}}

x and y are not différent, the initial hypothesis is false there the only solution is

x y ( x − y ) = 0 {\displaystyle xy(x-y)=0}

The proofs : we have

y i 2 − x i 2 = ( y − x ) ( x i + y i ) = ( y i + 1 − x i + 1 ) ( x i + 1 + y i + 1 + 2 x i + 1 y i + 1 ) {\displaystyle y_{i}^{2}-x_{i}^{2}=(y-x)(x_{i}+y_{i})=(y_{i+1}-x_{i+1})(x_{i+1}+y_{i+1}+2{\sqrt {x_{i+1}y_{i+1}}})}

= y i + 1 2 − x i + 1 2 + 2 x i + 1 y i + 1 ( y i + 1 − x i + 1 ) {\displaystyle =y_{i+1}^{2}-x_{i+1}^{2}+2{\sqrt {x_{i+1}y_{i+1}}}(y_{i+1}-x_{i+1})}

y i + 1 = Z {\displaystyle {\sqrt {y_{i+1}}}=Z} is solution of the following equation

Z 4 + 2 x i + 1 Z 3 − 2 x i + 1 3 Z − x i + 1 2 + x i 2 − y i 2 = 0 {\displaystyle Z^{4}+2{\sqrt {x_{i+1}}}Z^{3}-2{\sqrt {x_{i+1}^{3}}}Z-x_{i+1}^{2}+x_{i}^{2}-y_{i}^{2}=0}

Also  x i + 1 = Z ′ {\displaystyle {\sqrt {x_{i+1}}}=Z'} is solution of

Z ′ 4 + 2 y i + 1 Z ′ 3 − 2 y i + 1 3 Z ′ − y i + 1 2 + y i 2 − x i 2 = 0 {\displaystyle {Z'}^{4}+2{\sqrt {y_{i+1}}}{Z'}^{3}-2{\sqrt {y_{i+1}^{3}}}Z'-y_{i+1}^{2}+y_{i}^{2}-x_{i}^{2}=0}

And

Z 4 + 2 Z ′ Z 3 − 2 Z ′ 3 Z − Z ′ 4 + x i 2 − y i 2 = 0 {\displaystyle Z^{4}+2Z'Z^{3}-2{Z'}^{3}Z-{Z'}^{4}+x_{i}^{2}-y_{i}^{2}=0}

Also

Z ′ 4 + 2 Z Z ′ 3 − 2 Z 3 Z ′ − Z 4 + y i 2 − x i 2 = 0 {\displaystyle {Z'}^{4}+2Z{Z'}^{3}-2Z^{3}Z'-Z^{4}+y_{i}^{2}-x_{i}^{2}=0}

Let

Z = u − Z ′ 2 {\displaystyle Z=u-{\frac {Z'}{2}}}

But

( u − Z ′ 2 ) 4 + 2 Z ′ ( u − Z ′ 2 ) 3 − 2 Z ′ 3 ( u − Z ′ 2 ) − Z ′ 4 + x i 2 − y i 2 = 0 {\displaystyle (u-{\frac {Z'}{2}})^{4}+2Z'(u-{\frac {Z'}{2}})^{3}-2{Z'}^{3}(u-{\frac {Z'}{2}})-{Z'}^{4}+x_{i}^{2}-y_{i}^{2}=0}

Hence

u 4 + Z ′ 4 16 − 2 u 3 Z ′ + 3 2 u 2 Z ′ 2 + {\displaystyle u^{4}+{\frac {{Z'}^{4}}{16}}-2u^{3}{Z'}+{\frac {3}{2}}u^{2}{Z'}^{2}+}

− 1 2 u Z ′ 3 + 2 Z ′ u 3 − 3 u 2 Z ′ 2 + 3 2 u Z ′ 3 − 1 4 Z ′ 4 + {\displaystyle -{\frac {1}{2}}u{Z'}^{3}+2Z'u^{3}-3u^{2}{Z'}^{2}+{\frac {3}{2}}u{Z'}^{3}-{\frac {1}{4}}{Z'}^{4}+}

− 2 Z ′ 3 u + Z ′ 4 − Z ′ 4 + x i 2 − y i 2 = 0 {\displaystyle -2{Z'}^{3}u+{Z'}^{4}-{Z'}^{4}+x_{i}^{2}-y_{i}^{2}=0}

= u 4 − ( 3 2 Z ′ 2 ) u 2 − Z ′ 3 u − 3 16 Z ′ 4 + x i 2 − y i 2 = 0 {\displaystyle =u^{4}-({\frac {3}{2}}{Z'}^{2})u^{2}-{Z'}^{3}u-{\frac {3}{16}}{Z'}^{4}+x_{i}^{2}-y_{i}^{2}=0}

We deduce

( u 2 − 3 4 Z ′ 2 ) 2 − 3 4 Z ′ 4 − Z ′ 3 u + x i 2 − y i 2 = 0 {\displaystyle (u^{2}-{\frac {3}{4}}{Z'}^{2})^{2}-{\frac {3}{4}}{Z'}^{4}-{Z'}^{3}u+x_{i}^{2}-y_{i}^{2}=0}

Also  Z ′ = v − Z 2 {\displaystyle Z'=v-{\frac {Z}{2}}} leads to

( v − Z 2 ) 4 + 2 Z ( v − Z 2 ) 3 − 2 Z 3 ( v − Z 2 ) − Z 4 + y i 2 − x i 2 = 0 {\displaystyle (v-{\frac {Z}{2}})^{4}+2Z(v-{\frac {Z}{2}})^{3}-2{Z}^{3}(v-{\frac {Z}{2}})-{Z}^{4}+y_{i}^{2}-x_{i}^{2}=0}

Hence

v 4 + Z 4 16 − 2 v 3 Z + 3 2 v 2 Z 2 + {\displaystyle v^{4}+{\frac {{Z}^{4}}{16}}-2v^{3}{Z}+{\frac {3}{2}}v^{2}{Z}^{2}+}

− 1 2 v Z 3 + 2 Z v 3 − 3 v 2 Z 2 + 3 2 v Z 3 − 1 4 Z 4 + {\displaystyle -{\frac {1}{2}}v{Z}^{3}+2Zv^{3}-3v^{2}{Z}^{2}+{\frac {3}{2}}v{Z}^{3}-{\frac {1}{4}}{Z}^{4}+}

− 2 Z 3 v + Z 4 − Z 4 + y i 2 − x i 2 = 0 {\displaystyle -2{Z}^{3}v+{Z}^{4}-{Z}^{4}+y_{i}^{2}-x_{i}^{2}=0}

= v 4 − ( 3 2 Z 2 ) v 2 − Z 3 v − 3 16 Z 4 + y i 2 − x i 2 = 0 {\displaystyle =v^{4}-({\frac {3}{2}}{Z}^{2})v^{2}-{Z}^{3}v-{\frac {3}{16}}{Z}^{4}+y_{i}^{2}-x_{i}^{2}=0}

We deduce

( v 2 − 3 4 Z 2 ) 2 − 3 4 Z 4 − Z 3 v + y i 2 − x i 2 = 0 {\displaystyle (v^{2}-{\frac {3}{4}}{Z}^{2})^{2}-{\frac {3}{4}}{Z}^{4}-{Z}^{3}v+y_{i}^{2}-x_{i}^{2}=0}

If we add

( u 2 − 3 4 Z ′ 2 ) 2 + ( v 2 − 3 4 Z 2 ) 2 − 3 4 ( Z 4 + Z ′ 4 ) − Z ′ 3 u − Z 3 v = 0 {\displaystyle (u^{2}-{\frac {3}{4}}{Z'}^{2})^{2}+(v^{2}-{\frac {3}{4}}Z^{2})^{2}-{\frac {3}{4}}(Z^{4}+{Z'}^{4})-{Z'}^{3}u-Z^{3}v=0}

= ( Z 2 + Z ′ 2 4 + Z Z ′ − 3 4 Z 2 ) 2 + ( Z ′ 2 + Z 2 4 + Z ′ Z − 3 4 Z ′ 2 ) 2 − 3 4 ( Z 4 + Z ′ 4 ) − Z Z ′ 3 − Z 3 Z ′ − Z ′ 4 + Z 4 2 = 0 {\displaystyle =(Z^{2}+{\frac {{Z'}^{2}}{4}}+ZZ'-{\frac {3}{4}}Z^{2})^{2}+({Z'}^{2}+{\frac {Z^{2}}{4}}+Z'Z-{\frac {3}{4}}{Z'}^{2})^{2}-{\frac {3}{4}}(Z^{4}+{Z'}^{4})-Z{Z'}^{3}-Z^{3}Z'-{\frac {{Z'}^{4}+Z^{4}}{2}}=0}

= ( Z 2 + Z ′ 2 4 + Z Z ′ ) 2 + ( Z 2 + Z ′ 2 4 + Z ′ Z ) 2 − 5 4 ( Z 4 + Z ′ 4 ) − Z Z ′ ( Z 2 + Z ′ 2 ) = 0 {\displaystyle =({\frac {Z^{2}+{Z'}^{2}}{4}}+ZZ')^{2}+({\frac {Z^{2}+{Z'}^{2}}{4}}+Z'Z)^{2}-{\frac {5}{4}}(Z^{4}+{Z'}^{4})-ZZ'(Z^{2}+{Z'}^{2})=0}

= Z 4 + Z ′ 4 + 2 Z 2 Z ′ 2 8 + 2 Z 2 Z ′ 2 + Z Z ′ ( Z 2 + Z ′ 2 ) − 5 4 ( Z 4 + Z ′ 4 ) − Z Z ′ ( Z 2 + Z ′ 2 ) = 0 {\displaystyle ={\frac {Z^{4}+{Z'}^{4}+2Z^{2}{Z'}^{2}}{8}}+2Z^{2}{Z'}^{2}+ZZ'(Z^{2}+{Z'}^{2})-{\frac {5}{4}}(Z^{4}+{Z'}^{4})-ZZ'(Z^{2}+{Z'}^{2})=0}

= − 9 8 ( Z 4 + Z ′ 4 ) + 9 4 Z 2 Z ′ 2 = 0 {\displaystyle =-{\frac {9}{8}}(Z^{4}+{Z'}^{4})+{\frac {9}{4}}Z^{2}{Z'}^{2}=0}

= − ( Z 2 − Z ′ 2 ) 9 8 = 0 {\displaystyle =-(Z^{2}-{Z'}^{2}){\frac {9}{8}}=0}

The solution is

Z 2 − Z ′ 2 = y − x = 0 {\displaystyle Z^{2}-{Z'}^{2}=y-x=0}

Another proof : we have

x i − 1 + x i = x i − 1 − x i x i − 1 − x i = x i y i x i − 1 − x i {\displaystyle {\sqrt {x_{i-1}}}+{\sqrt {x_{i}}}={\frac {x_{i-1}-x_{i}}{{\sqrt {x_{i-1}}}-{\sqrt {x_{i}}}}}={\frac {\sqrt {x_{i}y_{i}}}{{\sqrt {x_{i-1}}}-{\sqrt {x_{i}}}}}}

= 1 x i − 1 x i y i − x i x i y i = 1 x i − 1 + y i − 1 x i y i − 1 y i {\displaystyle ={\frac {1}{{\sqrt {\frac {x_{i-1}}{x_{i}y_{i}}}}-{\sqrt {\frac {x_{i}}{x_{i}y_{i}}}}}}={\frac {1}{{\sqrt {\frac {\sqrt {x_{i-1}+y_{i-1}}}{{\sqrt {x_{i}}}y_{i}}}}-{\frac {1}{\sqrt {y_{i}}}}}}}

= 1 x i − 1 + y i − 1 4 − x i 4 x i 4 y i = x i 4 y i x i − 1 + y i − 1 4 − x i − 1 x i − 1 + y i − 1 4 {\displaystyle ={\frac {1}{\frac {{\sqrt[{4}]{x_{i-1}+y_{i-1}}}-{\sqrt[{4}]{x_{i}}}}{{\sqrt[{4}]{x_{i}}}{\sqrt {y_{i}}}}}}={\frac {{\sqrt[{4}]{x_{i}}}{\sqrt {y_{i}}}}{{\sqrt[{4}]{x_{i-1}+y_{i-1}}}-{\frac {\sqrt {x_{i-1}}}{\sqrt[{4}]{x_{i-1}+y_{i-1}}}}}}}

= x i 4 y i ( x i − 1 + y i − 1 4 ) x i − 1 + y i − 1 − x i − 1 ≤ x i 4 y i x i − 1 + y i − 1 4 y i + 1 4 {\displaystyle ={\frac {{\sqrt[{4}]{x_{i}}}{\sqrt {y_{i}}}({\sqrt[{4}]{x_{i-1}+y_{i-1}}})}{{\sqrt {x_{i-1}+y_{i-1}}}-{\sqrt {x_{i-1}}}}}\leq {\frac {{\sqrt[{4}]{x_{i}}}{\sqrt {y_{i}}}{\sqrt[{4}]{x_{i-1}+y_{i-1}}}}{\sqrt[{4}]{y_{i+1}}}}}

Because

x i − 1 + y i − 1 − x i − 1 = y i − 1 x i − 1 + y i − 1 + x i − 1 ≥ y i + 1 4 {\displaystyle {\sqrt {x_{i-1}+y_{i-1}}}-{\sqrt {x_{i-1}}}={\frac {y_{i-1}}{{\sqrt {x_{i-1}+y_{i-1}}}+{\sqrt {x_{i-1}}}}}\geq {\sqrt[{4}]{y_{i+1}}}}

y i − 1 = y i x i − 1 + y i − 1 ≥ y i + 1 4 ( x i − 1 + y i − 1 + x i − 1 ) = y i x i + y i 4 ( x i − 1 + y i − 1 + x i − 1 ) {\displaystyle y_{i-1}={\sqrt {y_{i}}}{\sqrt {x_{i-1}+y_{i-1}}}\geq {{\sqrt[{4}]{y_{i+1}}}({\sqrt {x_{i-1}+y_{i-1}}}+{\sqrt {x_{i-1}}})}={\frac {\sqrt {y_{i}}}{\sqrt[{4}]{x_{i}+y_{i}}}}({\sqrt {x_{i-1}+y_{i-1}}}+{\sqrt {x_{i-1}}})}

And

x i + y i 4 x i − 1 + y i − 1 ≥ x i − 1 + y i − 1 + x i − 1 {\displaystyle {\sqrt[{4}]{x_{i}+y_{i}}}{\sqrt {x_{i-1}+y_{i-1}}}\geq {{\sqrt {x_{i-1}+y_{i-1}}}+{\sqrt {x_{i-1}}}}}

Because

x i + y i 4 = x i − 1 2 + y i − 1 2 4 x i − 1 + y i − 1 4 ≥ 1 + x i − 1 x i − 1 + y i − 1 {\displaystyle {\sqrt[{4}]{x_{i}+y_{i}}}={\frac {\sqrt[{4}]{x_{i-1}^{2}+y_{i-1}^{2}}}{\sqrt[{4}]{x_{i-1}+y_{i-1}}}}\geq {1+{\sqrt {\frac {x_{i-1}}{x_{i-1}+y_{i-1}}}}}}

And

x i − 1 2 + y i − 1 2 4 ≥ x i − 1 + y i − 1 4 + x i − 1 + y i − 1 4 x i − 1 x i − 1 + y i − 1 {\displaystyle {\sqrt[{4}]{x_{i-1}^{2}+y_{i-1}^{2}}}\geq {{\sqrt[{4}]{x_{i-1}+y_{i-1}}}+{\sqrt[{4}]{x_{i-1}+y_{i-1}}}{\frac {\sqrt {x_{i-1}}}{\sqrt {x_{i-1}+y_{i-1}}}}}}

We deduce

x i − 1 + x i ≤ x i 4 y i y i + 1 4 = x i 4 y i + 1 4 ( x i + y i 4 ) y i + 1 4 = x i 4 ( x i + y i 4 ) {\displaystyle {\sqrt {x_{i-1}}}+{\sqrt {x_{i}}}\leq {\frac {{\sqrt[{4}]{x_{i}}}{\sqrt {y_{i}}}}{\sqrt[{4}]{y_{i+1}}}}={\frac {{\sqrt[{4}]{x_{i}}}{\sqrt[{4}]{y_{i+1}}}({\sqrt[{4}]{x_{i}+y_{i}}})}{\sqrt[{4}]{y_{i+1}}}}={\sqrt[{4}]{x_{i}}}({\sqrt[{4}]{x_{i}+y_{i}}})}

In the infinity

0 < x − y ≤ lim i ⟶ ∞ ( x i − 1 + x i ) = 2 x − y ≤ lim i ⟶ ∞ ( x i 4 x i + y i 4 ) = x − y {\displaystyle 0<{\sqrt {x-y}}\leq {\lim _{i\longrightarrow {\infty }}{({\sqrt {x_{i-1}}}+{\sqrt {x_{i}}})}=2{\sqrt {x-y}}}\leq {\lim _{i\longrightarrow {\infty }}{({\sqrt[{4}]{x_{i}}}{\sqrt[{4}]{x_{i}+y_{i}}})}={\sqrt {x-y}}}}

Therefore

x − y = 2 x − y = 0 {\displaystyle {\sqrt {x-y}}=2{\sqrt {x-y}}=0}

Another proof : let

x i = a i x i + 1 = x i + 1 + x i + 1 y i + 1 ⇒ a i = 1 + y i + 1 x i + 1 = 1 + y i x i {\displaystyle x_{i}=a_{i}x_{i+1}=x_{i+1}+{\sqrt {x_{i+1}y_{i+1}}}\Rightarrow {a_{i}=1+{\sqrt {\frac {y_{i+1}}{x_{i+1}}}}=1+{\frac {y_{i}}{x_{i}}}}}

Also

y i = b i y i + 1 = y i + 1 + x i + 1 y i + 1 ⇒ b i = 1 + x i + 1 y i + 1 = 1 + x i y i {\displaystyle y_{i}=b_{i}y_{i+1}=y_{i+1}+{\sqrt {x_{i+1}y_{i+1}}}\Rightarrow {b_{i}=1+{\sqrt {\frac {x_{i+1}}{y_{i+1}}}}=1+{\frac {x_{i}}{y_{i}}}}}

We deduce

a i = 1 + a i b i = 1 + y i x i {\displaystyle a_{i}=1+{\frac {a_{i}}{b_{i}}}=1+{\frac {y_{i}}{x_{i}}}}

And

b i = 1 + b i a i = 1 + x i y i {\displaystyle b_{i}=1+{\frac {b_{i}}{a_{i}}}=1+{\frac {x_{i}}{y_{i}}}}

Thus

a i b i = a i + b i {\displaystyle a_{i}b_{i}=a_{i}+b_{i}}

And

x i = c i y i + 1 = x i + 1 + x i + 1 y i + 1 ⇒ c i = b i ( b i − 1 ) {\displaystyle x_{i}=c_{i}y_{i+1}=x_{i+1}+{\sqrt {x_{i+1}y_{i+1}}}\Rightarrow {c_{i}=b_{i}(b_{i}-1)}}

Also

y i = d i x i + 1 = y i + 1 + x i + 1 y i + 1 ⇒ d i = a i ( a i − 1 ) {\displaystyle y_{i}=d_{i}x_{i+1}=y_{i+1}+{\sqrt {x_{i+1}y_{i+1}}}\Rightarrow {d_{i}=a_{i}(a_{i}-1)}}

We have

x i = b i x i + 1 y i + 1 , y i = a i x i + 1 y i + 1 {\displaystyle x_{i}=b_{i}{\sqrt {x_{i+1}y_{i+1}}},\quad {y_{i}=a_{i}{\sqrt {x_{i+1}y_{i+1}}}}}

Let also

x i = i α i , y i = i β i {\displaystyle x_{i}=i{\alpha }_{i},\quad {y_{i}=i{\beta }_{i}}}

And

α i = a i ′ α i + 1 , β i = b i ′ β i + 1 {\displaystyle {\alpha }_{i}=a'_{i}{\alpha }_{i+1},\quad {{\beta }_{i}=b'_{i}{\beta }_{i+1}}}

We deduce

a i b i = y i x i = β i α i = a i ′ b i ′ {\displaystyle {\frac {a_{i}}{b_{i}}}={\frac {y_{i}}{x_{i}}}={\frac {{\beta }_{i}}{{\alpha }_{i}}}={\frac {a'_{i}}{b'_{i}}}}

Because

i α i = x i = x i + 1 + x i + 1 y i + 1 = ( i + 1 ) ( α i + 1 + α i + 1 β i + 1 ) {\displaystyle i{\alpha }_{i}=x_{i}=x_{i+1}+{\sqrt {x_{i+1}y_{i+1}}}=(i+1)({\alpha }_{i+1}+{\sqrt {{\alpha }_{i+1}{\beta }_{i+1}}})}

i β i = y i = y i + 1 + x i + 1 y i + 1 = ( i + 1 ) ( β i + 1 + α i + 1 β i + 1 ) {\displaystyle i{\beta }_{i}=y_{i}=y_{i+1}+{\sqrt {x_{i+1}y_{i+1}}}=(i+1)({\beta }_{i+1}+{\sqrt {{\alpha }_{i+1}{\beta }_{i+1}}})}

α i β i = α i + 1 β i + 1 = a i ′ α i + 1 b i ′ β i + 1 {\displaystyle {\frac {{\alpha }_{i}}{{\beta }_{i}}}={\sqrt {\frac {{\alpha }_{i+1}}{{\beta }_{i+1}}}}={\frac {a'_{i}{\alpha }_{i+1}}{b'_{i}{\beta }_{i+1}}}}

And

b i ′ = i + 1 i b i ≠ 2 , a i ′ = i + 1 i a i ≠ 2 {\displaystyle b'_{i}={\frac {i+1}{i}}b_{i}\neq {2},\quad {a'_{i}={\frac {i+1}{i}}a_{i}\neq {2}}}

But

b i ′ − 2 = α i α i + 1 β i + 1 − 2 = a i ′ α i β i − 2 {\displaystyle b'_{i}-2={\frac {{\alpha }_{i}}{\sqrt {{\alpha }_{i+1}{\beta }_{i+1}}}}-2=a'_{i}{\frac {{\alpha }_{i}}{{\beta }_{i}}}-2}

= ( a i ′ − 2 ) x i y i + 2 ( x i y i − 1 ) {\displaystyle =(a'_{i}-2){\frac {x_{i}}{y_{i}}}+2({\frac {x_{i}}{y_{i}}}-1)}

And

a i ′ − 2 = β i α i + 1 β i + 1 − 2 = b i ′ β i α i − 2 {\displaystyle a'_{i}-2={\frac {{\beta }_{i}}{\sqrt {{\alpha }_{i+1}{\beta }_{i+1}}}}-2=b'_{i}{\frac {{\beta }_{i}}{{\alpha }_{i}}}-2}

= ( b i ′ − 2 ) y i x i + 2 ( y i x i − 1 ) {\displaystyle =(b'_{i}-2){\frac {y_{i}}{x_{i}}}+2({\frac {y_{i}}{x_{i}}}-1)}

Thus

( a i ′ − 2 ) 2 x i y i + 2 ( x i y i − 1 ) ( a i ′ − 2 ) {\displaystyle (a'_{i}-2)^{2}{\frac {x_{i}}{y_{i}}}+2({\frac {x_{i}}{y_{i}}}-1)(a'_{i}-2)}

= ( b i ′ − 2 ) 2 y i x i + 2 ( y i x i − 1 ) ( b i ′ − 2 ) {\displaystyle =(b'_{i}-2)^{2}{\frac {y_{i}}{x_{i}}}+2({\frac {y_{i}}{x_{i}}}-1)(b'_{i}-2)}

= ( ( a i ′ − 2 ) x i y i + 2 ( x i y i − 1 ) ) 2 y i x i + 2 ( y i x i − 1 ) ( ( a i ′ − 2 ) x i y i + 2 ( y i x i − 1 ) ) {\displaystyle =((a'_{i}-2){\frac {x_{i}}{y_{i}}}+2({\frac {x_{i}}{y_{i}}}-1))^{2}{\frac {y_{i}}{x_{i}}}+2({\frac {y_{i}}{x_{i}}}-1)((a'_{i}-2){\frac {x_{i}}{y_{i}}}+2({\frac {y_{i}}{x_{i}}}-1))}

= ( a i ′ − 2 ) 2 x i y i + 4 ( x i y i − 1 ) 2 y i x i + 4 ( x i y i − 1 ) ( a i ′ − 2 ) + 2 ( 1 − x i y i ) ( a i ′ − 2 ) + 4 ( y i x i − 1 ) 2 {\displaystyle =(a'_{i}-2)^{2}{\frac {x_{i}}{y_{i}}}+4({\frac {x_{i}}{y_{i}}}-1)^{2}{\frac {y_{i}}{x_{i}}}+4({\frac {x_{i}}{y_{i}}}-1)(a'_{i}-2)+2(1-{\frac {x_{i}}{y_{i}}})(a'_{i}-2)+4({\frac {y_{i}}{x_{i}}}-1)^{2}}

= ( a i ′ − 2 ) 2 x i y i + 4 ( x i y i − 1 ) 2 y i x i + 2 ( x i y i − 1 ) ( a i ′ − 2 ) + 4 ( y i x i − 1 ) 2 {\displaystyle =(a'_{i}-2)^{2}{\frac {x_{i}}{y_{i}}}+4({\frac {x_{i}}{y_{i}}}-1)^{2}{\frac {y_{i}}{x_{i}}}+2({\frac {x_{i}}{y_{i}}}-1)(a'_{i}-2)+4({\frac {y_{i}}{x_{i}}}-1)^{2}}

Hence

4 ( x i y i − 1 ) 2 y i x i + 4 ( y i x i − 1 ) 2 = 0 {\displaystyle 4({\frac {x_{i}}{y_{i}}}-1)^{2}{\frac {y_{i}}{x_{i}}}+4({\frac {y_{i}}{x_{i}}}-1)^{2}=0}

= 4 ( ( x − y ) 2 y i 2 ) y i x i + 4 ( ( x − y ) 2 x i 2 ) = 0 {\displaystyle =4({\frac {(x-y)^{2}}{y_{i}^{2}}}){\frac {y_{i}}{x_{i}}}+4({\frac {(x-y)^{2}}{x_{i}^{2}}})=0}

= 4 ( x − y ) 2 ( 1 x i y i + 1 x i 2 ) = 0 {\displaystyle =4(x-y)^{2}({\frac {1}{x_{i}y_{i}}}+{\frac {1}{x_{i}^{2}}})=0}

⇒ ( x − y ) 2 = 0 {\displaystyle \Rightarrow {(x-y)^{2}=0}}

Another proof : We have

c i d i = a i b i {\displaystyle c_{i}d_{i}=a_{i}b_{i}}

Let

a i − 1 b i − 1 = f i a i b i {\displaystyle a_{i-1}b_{i-1}=f_{i}a_{i}b_{i}}

We have

f i = a i − 1 b i − 1 a i b i = x i − 1 y i − 1 x i + 1 y i + 1 x i 2 y i 2 {\displaystyle f_{i}={\frac {a_{i-1}b_{i-1}}{a_{i}b_{i}}}={\frac {x_{i-1}y_{i-1}x_{i+1}y_{i+1}}{x_{i}^{2}y_{i}^{2}}}}

= x i y i ( x i + y i ) 2 x i + 1 y i + 1 x i + 1 y i + 1 ( x i + y i ) 2 ≤ 1 {\displaystyle ={\frac {{\sqrt {x_{i}y_{i}}}({\sqrt {x_{i}}}+{\sqrt {y_{i}}})^{2}x_{i+1}y_{i+1}}{x_{i+1}y_{i+1}(x_{i}+y_{i})^{2}}}\leq {1}}

a i b i = a i b i + b i a i + 2 = ( a i b i + b i a i ) 2 = ( a i − 1 b i − 1 − 2 ) 2 {\displaystyle a_{i}b_{i}={\frac {a_{i}}{b_{i}}}+{\frac {b_{i}}{a_{i}}}+2=({\sqrt {\frac {a_{i}}{b_{i}}}}+{\sqrt {\frac {b_{i}}{a_{i}}}})^{2}=(a_{i-1}b_{i-1}-2)^{2}}

And

a i b i = e i = ( f i a i b i − 2 ) 2 {\displaystyle a_{i}b_{i}=e_{i}=(f_{i}a_{i}b_{i}-2)^{2}}

It means

e i 2 f i 2 − ( 4 f i + 1 ) e i + 4 = 0 {\displaystyle e_{i}^{2}f_{i}^{2}-(4f_{i}+1)e_{i}+4=0}

And

e i = a i b i = 4 f i + 1 + 8 f i + 1 2 f i 2 {\displaystyle e_{i}=a_{i}b_{i}={\frac {4f_{i}+1+{\sqrt {8f_{i}+1}}}{2f_{i}^{2}}}}

But

a i b i = f i a i b i − 2 ⇒ f i a i b i − a i b i − 2 = 0 {\displaystyle {\sqrt {a_{i}b_{i}}}=f_{i}a_{i}b_{i}-2\Rightarrow {f_{i}a_{i}b_{i}-{\sqrt {a_{i}b_{i}}}-2=0}}

Thus

e i = a i b i = 1 + 1 + 8 f i 2 f i {\displaystyle {\sqrt {e_{i}}}={\sqrt {a_{i}b_{i}}}={\frac {1+{\sqrt {1+8f_{i}}}}{2f_{i}}}}

And

e i − 2 = 1 − f i + 1 + 8 f i − 3 f i 2 f i = 1 − f i + 1 − f i 2 + 8 f i ( 1 − f i ) 1 + 8 f i + 3 f i 2 f i {\displaystyle {\sqrt {e_{i}}}-2={\frac {1-f_{i}+{\sqrt {1+8f_{i}}}-3f_{i}}{2f_{i}}}={\frac {1-f_{i}+{\frac {1-f_{i}^{2}+8f_{i}(1-f_{i})}{{\sqrt {1+8f_{i}}}+3f_{i}}}}{2f_{i}}}}

= ( 1 − f i ) 1 + 1 + 9 f i 8 f i 1 + 3 f i 2 f i {\displaystyle =(1-f_{i}){\frac {1+{\frac {1+9f_{i}}{{\sqrt {8f_{i}1}}+3f_{i}}}}{2f_{i}}}}

And

e i − 4 = a i b i − 4 = ( e i − 2 ) ( e i + 2 ) = ( e i − 2 ) f i a i b i = ( 1 − f i ) f i a i b i {\displaystyle e_{i}-4=a_{i}b_{i}-4=({\sqrt {e_{i}}}-2)({\sqrt {e_{i}}}+2)=({\sqrt {e_{i}}}-2)f_{i}a_{i}b_{i}=(1-f_{i})f_{i}a_{i}b_{i}}

1 − f i 2 + 4 f i ( 1 − f i ) + 1 − f i 2 + 8 f i ( 1 − f i ) 8 f i + 1 + 3 f i 2 2 f i 2 {\displaystyle {\frac {1-f_{i}^{2}+4f_{i}(1-f_{i})+{\frac {1-f_{i}^{2}+8f_{i}(1-f_{i})}{{\sqrt {8f_{i}+1}}+3f_{i}^{2}}}}{2f_{i}^{2}}}}

= ( 1 − f i ) 1 + 5 f i + 1 + 9 f i 8 f i + 1 + 3 f i 2 2 f i 2 {\displaystyle =(1-f_{i}){\frac {1+5f_{i}+{\frac {1+9f_{i}}{{\sqrt {8f_{i}+1}}+3f_{i}^{2}}}}{2f_{i}^{2}}}}

( 1 − f i ) ( f i a i b i − 1 + 5 f i + 1 + 9 f i 1 + 8 f i + 3 f i 2 2 f i 2 ) = 0 {\displaystyle (1-f_{i})(f_{i}a_{i}b_{i}-{\frac {1+5f_{i}+{\frac {1+9f_{i}}{{\sqrt {1+8f_{i}}}+3f_{i}^{2}}}}{2f_{i}^{2}}})=0}

( 1 − f i ) ( a i b i − 1 + 5 f i + 1 + 9 f i 1 + f i + 3 f i 2 2 f i 3 ) = 0 {\displaystyle (1-f_{i})(a_{i}b_{i}-{\frac {1+5f_{i}+{\frac {1+9f_{i}}{{\sqrt {1+f_{i}}}+3f_{i}^{2}}}}{2f_{i}^{3}}})=0}

= ( 1 − f i ) ( f i ( 1 + 4 f i ) + f i 8 f i + 1 − 1 − 5 f i − 1 + 9 f i 1 + 8 f i + 3 f i 2 2 f i 3 ) = 0 {\displaystyle =(1-f_{i})({\frac {f_{i}(1+4f_{i})+f_{i}{\sqrt {8f_{i}+1}}-1-5f_{i}-{\frac {1+9f_{i}}{{\sqrt {1+8f_{i}}}+3f_{i}^{2}}}}{2f_{i}^{3}}})=0}

The expression between the parenthesis is not equal to zero, we deduce

f i = 1 {\displaystyle f_{i}=1}

Else

a i b i = 2 f i a i b i − 1 {\displaystyle {\sqrt {a_{i}b_{i}}}={\frac {2}{f_{i}{\sqrt {a_{i}b_{i}}}-1}}}

a i b i − 2 = 2 ( 2 − f i a i b i f i a i b i ) = 2 ( 2 ( 1 − f i ) + f i ( 2 − a i b i ) f i a i b i ) {\displaystyle {\sqrt {a_{i}b_{i}}}-2=2({\frac {2-f_{i}{\sqrt {a_{i}b_{i}}}}{f_{i}{\sqrt {a_{i}b_{i}}}}})=2({\frac {2(1-f_{i})+f_{i}(2-{\sqrt {a_{i}b_{i}}})}{f_{i}{\sqrt {a_{i}b_{i}}}}})}

We deduce

( a i b i − 2 ) ( 1 + 2 a i b i ) = ( a i b i − 2 ) ( a i b i + 2 a i b i ) {\displaystyle ({\sqrt {a_{i}b_{i}}}-2)(1+{\frac {2}{\sqrt {a_{i}b_{i}}}})=({\sqrt {a_{i}b_{i}}}-2)({\frac {{\sqrt {a_{i}b_{i}}}+2}{\sqrt {a_{i}b_{i}}}})}

= ( a i b i − 2 ) ( f i a i b i a i b i ) = ( a i b i − 2 ) f i a i b i = 4 1 − f i f i a i b i {\displaystyle =({\sqrt {a_{i}b_{i}}}-2)({\frac {f_{i}a_{i}b_{i}}{\sqrt {a_{i}b_{i}}}})=({\sqrt {a_{i}b_{i}}}-2)f_{i}{\sqrt {a_{i}b_{i}}}=4{\frac {1-f_{i}}{f_{i}{\sqrt {a_{i}b_{i}}}}}}

Thus

( 1 + 1 + 8 f i 2 f i − 2 ) f i a i b i = ( 1 − f i ) ( 1 + 1 + 9 f i 1 + 8 f i + 3 f i 2 f i ) = ( 1 − f i ) 4 f i a i b i {\displaystyle ({\frac {1+{\sqrt {1+8f_{i}}}}{2f_{i}}}-2)f_{i}{\sqrt {a_{i}b_{i}}}=(1-f_{i})({\frac {1+{\frac {1+9f_{i}}{{\sqrt {1+8f_{i}}}+3f_{i}}}}{2f_{i}}})=(1-f_{i}){\frac {4}{f_{i}{\sqrt {a_{i}b_{i}}}}}}

The expressions between the parenthesis are not equal, therefore

f i = 1 {\displaystyle f_{i}=1}

Else

a i − 4 = ( e i − 2 ) ( e i + 2 ) = ( 1 − f i ) ( 1 + 1 + 9 f i 1 + 8 f i + 3 f i 2 f i ) ( 1 + 8 f i + 1 + 8 f i 2 f i ) {\displaystyle a_{i}-4=({\sqrt {e_{i}}}-2)({\sqrt {e_{i}}}+2)=(1-f_{i})({\frac {1+{\frac {1+9f_{i}}{{\sqrt {1+8f_{i}}}+3f_{i}}}}{2f_{i}}})({\frac {1+8f_{i}+{\sqrt {1+8f_{i}}}}{2f_{i}}})}

Hence

( 1 − f i ) ( 1 + 5 f i + 1 + 9 f i 1 + 8 f i + 3 f i 2 2 f i 2 − ( 1 + 1 + 9 f i 1 + 8 f i + 3 f i 2 f i ) ( 1 + 8 f i + 8 f i + 1 2 f i ) ) = 0 {\displaystyle (1-f_{i})({\frac {1+5f_{i}+{\frac {1+9f_{i}}{{\sqrt {1+8f_{i}}}+3f_{i}^{2}}}}{2f_{i}^{2}}}-({\frac {1+{\frac {1+9f_{i}}{{\sqrt {1+8f_{i}}}+3f_{i}}}}{2f_{i}}})({\frac {1+8f_{i}+{\sqrt {8f_{i}+1}}}{2f_{i}}}))=0}

( 1 − f i ) ( 2 + 10 f i + 2 1 + 9 f i 1 + 8 f i + 3 f i 2 − ( 1 + 8 f i + ( 1 + 9 f i ) 8 f i + 1 8 f i + + 3 f i + 8 f i + 1 + ( 1 + 8 f i ) ( 1 + 9 f i ) 8 f i + 1 + 3 f i ) ) = 0 {\displaystyle (1-f_{i})(2+10f_{i}+2{\frac {1+9f_{i}}{{\sqrt {1+8f_{i}}}+3f_{i}^{2}}}-(1+8f_{i}+{\frac {(1+9f_{i}){\sqrt {8f_{i}+1}}}{{\sqrt {8f_{i}+}}+3f_{i}}}+{\sqrt {8f_{i}+1}}+{\frac {(1+8f_{i})(1+9f_{i})}{{\sqrt {8f_{i}+1}}+3f_{i}}}))=0}

The expression between the parenthesis is not equal to zero, we deduce

f i = 1 ⇒ x i y i ( a i − b i − 4 ) = ( x − y ) 2 = 0 {\displaystyle f_{i}=1\Rightarrow {x_{i}y_{i}(a_{i}-b_{i}-4)=(x-y)^{2}=0}}

And, else

2 f i 2 e i = 4 f i + 1 + 8 f i + 1 {\displaystyle 2f_{i}^{2}e_{i}=4f_{i}+1+{\sqrt {8f_{i}+1}}}

( 2 f i 2 e i − 4 f i − 1 ) 2 = 8 f i + 1 = 4 f i 4 e i 2 + 16 f i 2 − 16 f i 3 e i − 4 f i 2 e i + 8 f i + 1 {\displaystyle (2f_{i}^{2}e_{i}-4f_{i}-1)^{2}=8f_{i}+1=4f_{i}^{4}e_{i}^{2}+16f_{i}^{2}-16f_{i}^{3}e_{i}-4f_{i}^{2}e_{i}+8f_{i}+1}

f i 2 e i 2 + 4 − 4 f i e i − e i = 0 {\displaystyle f_{i}^{2}e_{i}^{2}+4-4f_{i}e_{i}-e_{i}=0}

( f i 2 e i − 4 f i − 1 ) e i = − 4 {\displaystyle (f_{i}^{2}e_{i}-4f_{i}-1)e_{i}=-4}

e i = 4 1 + 4 f i − f i 2 e i {\displaystyle e_{i}={\frac {4}{1+4f_{i}-f_{i}^{2}e_{i}}}}

e i − 4 = − 16 f i + 4 f i 2 e i 1 + 4 f i − f i 2 e i {\displaystyle e_{i}-4={\frac {-16f_{i}+4f_{i}^{2}e_{i}}{1+4f_{i}-f_{i}^{2}e_{i}}}}

= − 16 f i + 16 f i 2 + 4 f i 2 ( e i − 4 ) 1 + 4 f i − f i 2 e i {\displaystyle ={\frac {-16f_{i}+16f_{i}^{2}+4f_{i}^{2}(e_{i}-4)}{1+4f_{i}-f_{i}^{2}e_{i}}}}

( e i − 4 ) ( 1 − 4 f i 2 1 + 4 f i − f i 2 e i ) {\displaystyle (e_{i}-4)(1-{\frac {4f_{i}^{2}}{1+4f_{i}-f_{i}^{2}e_{i}}})}

= ( f i − 1 ) 16 f i 1 + 4 f i − f i 2 e i = 4 f i e i ( f i − 1 ) {\displaystyle =(f_{i}-1){\frac {16f_{i}}{1+4f_{i}-f_{i}^{2}e_{i}}}=4f_{i}e_{i}(f_{i}-1)}

= ( e i − 4 ) ( 1 − f i 2 e i ) {\displaystyle =(e_{i}-4)(1-f_{i}^{2}e_{i})}

e i = 4 f i + 1 + 8 f i + 1 2 f i 2 {\displaystyle e_{i}={\frac {4f_{i}+1+{\sqrt {8f_{i}+1}}}{2f_{i}^{2}}}}

e i − 4 = 4 f i ( 1 − f i ) + 1 − f i 2 + 8 f i + 1 − 3 f i 2 2 f i 2 {\displaystyle e_{i}-4={\frac {4f_{i}(1-f_{i})+1-f_{i}^{2}+{\sqrt {8f_{i}+1}}-3f_{i}^{2}}{2f_{i}^{2}}}}

= 4 f i ( 1 − f i ) + ( 1 − f i ) ( 1 + f i ) + 8 f i − 8 f i 4 + 1 − f i 4 8 f i + 1 + 3 f i 2 2 f i 2 {\displaystyle ={\frac {4f_{i}(1-f_{i})+(1-f_{i})(1+f_{i})+{\frac {8f_{i}-8f_{i}^{4}+1-f_{i}^{4}}{{\sqrt {8f_{i}+1}}+3f_{i}^{2}}}}{2f_{i}^{2}}}}

= ( 1 − f i ) ( 5 f i + 1 + 9 f i 3 + 9 f i 2 + 9 f i + 1 8 f i + 1 + 3 f i 2 2 f i 2 ) {\displaystyle =(1-f_{i})({\frac {5f_{i}+1+{\frac {9f_{i}^{3}+9f_{i}^{2}+9f_{i}+1}{{\sqrt {8f_{i}+1}}+3f_{i}^{2}}}}{2f_{i}^{2}}})}

( f i − 1 ) ( 4 f i e i + ( 1 − f i 2 e i ) ( 5 f i + 1 + 9 f i 3 + 9 f i 2 + 9 f i + 1 8 f i + 1 + 3 f i 2 2 f i 2 ) ) = 0 {\displaystyle (f_{i}-1)(4f_{i}e_{i}+(1-f_{i}^{2}e_{i})({\frac {5f_{i}+1+{\frac {9f_{i}^{3}+9f_{i}^{2}+9f_{i}+1}{{\sqrt {8f_{i}+1}}+3f_{i}^{2}}}}{2f_{i}^{2}}}))=0}

= ( f i − 1 ) ( 4 ( 4 f i + 1 + 8 f i + 1 ) 2 f i + ( 1 − 4 f i + 1 + 8 f i + 1 2 ) ( 5 f i + 1 + 9 f i 3 + 9 f i 2 + 9 f i + 1 8 f i + 1 + 3 f i 2 2 f i 2 ) ) = 0 {\displaystyle =(f_{i}-1)({\frac {4(4f_{i}+1+{\sqrt {8f_{i}+1}})}{2f_{i}}}+(1-{\frac {4f_{i}+1+{\sqrt {8f_{i}+1}}}{2}})({\frac {5f_{i}+1+{\frac {9f_{i}^{3}+9f_{i}^{2}+9f_{i}+1}{{\sqrt {8f_{i}+1}}+3f_{i}^{2}}}}{2f_{i}^{2}}}))=0}

= ( f i − 1 ) ( 5 f i + 1 + 9 f i 3 + 9 f i 2 + 9 f i + 1 8 f i + 1 + 3 f i 2 2 f i 2 − ( 4 f i + 1 + 8 f i + 1 2 ) ( − 3 f i + 1 + 9 f i 3 + 9 f i 2 + 9 f i + 1 8 f i + 1 + 3 f i 2 2 f i 2 ) ) = 0 {\displaystyle =(f_{i}-1)({\frac {5f_{i}+1+{\frac {9f_{i}^{3}+9f_{i}^{2}+9f_{i}+1}{{\sqrt {8f_{i}+1}}+3f_{i}^{2}}}}{2f_{i}^{2}}}-({\frac {4f_{i}+1+{\sqrt {8f_{i}+1}}}{2}})({\frac {-3f_{i}+1+{\frac {9f_{i}^{3}+9f_{i}^{2}+9f_{i}+1}{{\sqrt {8f_{i}+1}}+3f_{i}^{2}}}}{2f_{i}^{2}}}))=0}

And the expression between the parenthesis is not equal to zero, it means that

f i = 1 ⇒ a i b i = e i = 4 {\displaystyle f_{i}=1\Rightarrow {a_{i}b_{i}=e_{i}=4}}

a i 2 b i 2 − 4 a i b i = ( a i − b i ) 2 = 0 = ( y i − x i ) 2 x i + 1 y i + 1 {\displaystyle a_{i}^{2}b_{i}^{2}-4a_{i}b_{i}=(a_{i}-b_{i})^{2}=0=(y_{i}-x_{i})^{2}x_{i+1}y_{i+1}}

And

( a i − b i ) x i + 1 y i + 1 = y − x = 0 {\displaystyle (a_{i}-b_{i}){\sqrt {x_{i+1}y_{i+1}}}=y-x=0}

Another proof :

x i {\displaystyle x_{i}} and  y i {\displaystyle y_{i}} are particular cases of  v i {\displaystyle v_{i}} and  w i {\displaystyle w_{i}} which follow

v i = x i x i − y i ( x − y ) , w i = y i x i − y i ( x − y ) {\displaystyle v_{i}={\frac {x^{i}}{x^{i}-y^{i}}}(x-y),\quad {w_{i}={\frac {y^{i}}{x^{i}-y^{i}}}(x-y)}}

x i = v 2 i − 1 , y i = w 2 i − 1 {\displaystyle x_{i}=v_{2^{i-1}},\quad {y_{i}=w_{2^{i-1}}}}

Also

e i = v i + w i w i , f i = v i + w i v i {\displaystyle e_{i}={\frac {v_{i}+w_{i}}{w_{i}}},\quad {f_{i}={\frac {v_{i}+w_{i}}{v_{i}}}}}

e i = e i + f i f i , f i = e i + f i e i {\displaystyle e_{i}={\frac {e_{i}+f_{i}}{f_{i}}},\quad {f_{i}={\frac {e_{i}+f_{i}}{e_{i}}}}}

b i = e 2 i − 1 , a i = f 2 i − 1 {\displaystyle b_{i}=e_{2^{i-1}},\quad {a_{i}=f_{2^{i-1}}}}

So, we have

v i − w i = x − y {\displaystyle v_{i}-w_{i}=x-y}

And

e i f i = e i + f i {\displaystyle e_{i}f_{i}=e_{i}+f_{i}}

But

b 1 b 2 . . . b i = 1 + e 1 f 1 + . . . + e 2 i − 1 f 2 i − 1 {\displaystyle b_{1}b_{2}...b_{i}=1+{\frac {e_{1}}{f_{1}}}+...+{\frac {e_{2^{i}-1}}{f_{2^{i}-1}}}}

Let

1 + e 1 f 1 + e 2 f 2 + . . . + e i f i = 1 + v 1 w 1 + v 2 w 2 + . . . + v i w i {\displaystyle 1+{\frac {e_{1}}{f_{1}}}+{\frac {e_{2}}{f_{2}}}+...+{\frac {e_{i}}{f_{i}}}=1+{\frac {v_{1}}{w_{1}}}+{\frac {v_{2}}{w_{2}}}+...+{\frac {v_{i}}{w_{i}}}}

= 1 + x y + x 2 y 2 + . . . + x i y i = 1 − x ( i + 1 ) y ( i + 1 ) 1 − x y = 2 − e i + 1 2 − e 1 {\displaystyle =1+{\frac {x}{y}}+{\frac {x^{2}}{y^{2}}}+...+{\frac {x^{i}}{y^{i}}}={\frac {1-{\frac {x^{(i+1)}}{y^{(i+1)}}}}{1-{\frac {x}{y}}}}={\frac {2-e_{i+1}}{2-e_{1}}}}

Thus

e i + 1 − 2 = e i + 1 − f i + 1 f i + 1 = v i + 1 − w i + 1 w i + 1 = x − y w i + 1 {\displaystyle e_{i+1}-2={\frac {e_{i+1}-f_{i+1}}{f_{i+1}}}={\frac {v_{i+1}-w_{i+1}}{w_{i+1}}}={\frac {x-y}{w_{i+1}}}}

= ( e 1 − 2 ) ( 1 + ( e 1 − 1 ) + ( e 2 − 1 ) + . . . + ( e i − 1 ) ) = ( e 1 − 2 ) ( 1 + i + ( e 1 − 2 ) + ( e 2 − 2 ) + . . . + ( e i − 2 ) ) {\displaystyle =(e_{1}-2)(1+(e_{1}-1)+(e_{2}-1)+...+(e_{i}-1))=(e_{1}-2)(1+i+(e_{1}-2)+(e_{2}-2)+...+(e_{i}-2))}

= ( e 1 − f 1 f 1 ) ( 1 + i + e 1 − f 1 f 1 + e 2 − f 2 f 2 + . . . + e i − f i f i ) {\displaystyle =({\frac {e_{1}-f_{1}}{f_{1}}})(1+i+{\frac {e_{1}-f_{1}}{f_{1}}}+{\frac {e_{2}-f_{2}}{f_{2}}}+...+{\frac {e_{i}-f_{i}}{f_{i}}})}

= ( v 1 − w 1 w 1 ) ( 1 + i + v 1 − w 1 w 1 + v 2 − w 2 w 2 + . . . + v i − w i w i ) {\displaystyle =({\frac {v_{1}-w_{1}}{w_{1}}})(1+i+{\frac {v_{1}-w_{1}}{w_{1}}}+{\frac {v_{2}-w_{2}}{w_{2}}}+...+{\frac {v_{i}-w_{i}}{w_{i}}})}

( x − y ) ( 1 w i + 1 − 1 w 1 ( 1 + i + x − y w 1 + x − y w 2 + . . . + x − y w i ) ) = 0 {\displaystyle (x-y)({\frac {1}{w_{i+1}}}-{\frac {1}{w_{1}}}(1+i+{\frac {x-y}{w_{1}}}+{\frac {x-y}{w_{2}}}+...+{\frac {x-y}{w_{i}}}))=0}

And

1 w i + 1 − ( 1 + i ) 1 w 1 {\displaystyle {\frac {1}{w_{i+1}}}-(1+i){\frac {1}{w_{1}}}}

is not equal to zero for  x − y = 0 {\displaystyle x-y=0} , and it is the case, of course, of

− 1 w 1 ( x − y ) ( 1 w 1 + . . . + 1 w i ) {\displaystyle -{\frac {1}{w_{1}}}(x-y)({\frac {1}{w_{1}}}+...+{\frac {1}{w_{i}}})}

therefore the expression between the parenthesis is not equal to zero. We gave the fourth proof that the only solution is

x y ( x − y ) = 0 {\displaystyle xy(x-y)=0}

And there are others (see the series).

## The Ghanouchi's series

[edit | edit source]

As seen

x i y i = y i − 1 − y i = x i − 1 − x i {\displaystyle {\sqrt {x_{i}y_{i}}}=y_{i-1}-y_{i}=x_{i-1}-x_{i}}

we deduce

x i − x i + 1 = x i + 1 y i + 1 {\displaystyle x_{i}-x_{i+1}={\sqrt {x_{i+1}y_{i+1}}}}

x i − 1 − x i = x i y i {\displaystyle x_{i-1}-x_{i}={\sqrt {x_{i}y_{i}}}}

x 1 − x 2 = x − x 2 = x 2 y 2 {\displaystyle x_{1}-x_{2}=x-x_{2}={\sqrt {x_{2}y_{2}}}}

then

∑ j = 2 j = i + 1 ( x j y j ) = x − x 2 + x 2 − x 3 + . . . + x i − x i + 1 = x − x i + 1 {\displaystyle \sum _{j=2}^{j=i+1}{({\sqrt {x_{j}y_{j}}})}=x-x_{2}+x_{2}-x_{3}+...+x_{i}-x_{i+1}=x-x_{i+1}}

and

∑ j = 2 j = ∞ ( x j y j ) = lim i ⟶ ∞ ( x − x i + 1 ) {\displaystyle \sum _{j=2}^{j=\infty }{({\sqrt {x_{j}y_{j}}})}=\lim _{i\longrightarrow {\infty }}{(x-x_{i+1})}}

if  x > y {\displaystyle x>y}

∑ j = 2 j = ∞ ( x j y j ) = lim i ⟶ ∞ ( x − x i + 1 ) = x − ( x − y ) = y {\displaystyle \sum _{j=2}^{j=\infty }{({\sqrt {x_{j}y_{j}}})}=\lim _{i\longrightarrow {\infty }}{(x-x_{i+1})}=x-(x-y)=y}

if  x < y {\displaystyle x<y}

∑ j = 2 j = ∞ ( x j y j ) = lim i ⟶ ∞ ( x − x i + 1 ) = x {\displaystyle \sum _{j=2}^{j=\infty }{({\sqrt {x_{j}y_{j}}})}=\lim _{i\longrightarrow {\infty }}{(x-x_{i+1})}=x}

## The applications of the Ghanouchi's series

[edit | edit source]

∑ j = 2 j = i ( ( − 1 ) j x j y j ) = x − x 2 − ( x 2 − x 3 ) + ( x 3 − x 4 ) − . . . + ( − 1 ) i ( x i − 1 − x i ) {\displaystyle \sum _{j=2}^{j=i}{((-1)^{j}{\sqrt {x_{j}y_{j}}})}=x-x_{2}-(x_{2}-x_{3})+(x_{3}-x_{4})-...+(-1)^{i}(x_{i-1}-x_{i})}

= x − 2 x 2 + 2 x 3 − . . . + 2 ( − 1 ) i − 1 x i − 1 + ( − 1 ) i + 1 x i {\displaystyle =x-2x_{2}+2x_{3}-...+2(-1)^{i-1}x_{i-1}+(-1)^{i+1}x_{i}}

= 2 ∑ j = 2 j = i − 1 ( ( − 1 ) j + 1 x j ) + x + ( − 1 ) i + 1 x i {\displaystyle =2\sum _{j=2}^{j=i-1}{((-1)^{j+1}x_{j})}+x+(-1)^{i+1}x_{i}}

= 2 ∑ j = 1 j = i ( ( − 1 ) j + 1 x j ) − x − ( − 1 ) i + 1 x i {\displaystyle =2\sum _{j=1}^{j=i}{((-1)^{j+1}x_{j})}-x-(-1)^{i+1}x_{i}}

= ∑ j = 2 j = i − 1 ( ( − 1 ) j + 1 x j ) + ∑ j = 1 j = i ( ( − 1 ) j + 1 x j ) {\displaystyle =\sum _{j=2}^{j=i-1}{((-1)^{j+1}x_{j})}+\sum _{j=1}^{j=i}{((-1)^{j+1}x_{j})}}

= 2 ∑ j = 2 j = i − 1 ( ( − 1 ) j + 1 y j ) + y + ( − 1 ) i + 1 y i {\displaystyle =2\sum _{j=2}^{j=i-1}{((-1)^{j+1}y_{j})}+y+(-1)^{i+1}y_{i}}

= 2 ∑ j = 1 j = i ( ( − 1 ) j + 1 y j ) − y − ( − 1 ) i + 1 y i {\displaystyle =2\sum _{j=1}^{j=i}{((-1)^{j+1}y_{j})}-y-(-1)^{i+1}y_{i}}

= ∑ j = 2 j = i − 1 ( ( − 1 ) j + 1 y j ) + ∑ j = 1 j = i ( ( − 1 ) j + 1 y j ) {\displaystyle =\sum _{j=2}^{j=i-1}{((-1)^{j+1}y_{j})}+\sum _{j=1}^{j=i}{((-1)^{j+1}y_{j})}}

2 ∑ j = 1 j = i ( ( − 1 ) j + 1 x j ) = ∑ j = 2 j = i ( ( − 1 ) j x j y j ) + x + ( − 1 ) i + 1 x i {\displaystyle 2\sum _{j=1}^{j=i}{((-1)^{j+1}x_{j})}=\sum _{j=2}^{j=i}{((-1)^{j}{\sqrt {x_{j}y_{j}}})}+x+(-1)^{i+1}x_{i}}

and

2 ∑ j = 1 j = i ( ( − 1 ) j + 1 y j ) = ∑ j = 2 j = i ( ( − 1 ) j x j y j ) + y + ( − 1 ) i + 1 y i {\displaystyle 2\sum _{j=1}^{j=i}{((-1)^{j+1}y_{j})}=\sum _{j=2}^{j=i}{((-1)^{j}{\sqrt {x_{j}y_{j}}})}+y+(-1)^{i+1}y_{i}}

As we do not know the limit of  ( − 1 ) i + 1 x i {\displaystyle (-1)^{i+1}x_{i}} , then

∑ j = 1 j = ∞ ( ( − 1 ) j x j ) {\displaystyle \sum _{j=1}^{j=\infty }{((-1)^{j}x_{j})}}

can be not convergent. But

∑ j = 2 j = ∞ ( ( − 1 ) j x j y j ) {\displaystyle \sum _{j=2}^{j=\infty }{((-1)^{j}{\sqrt {x_{j}y_{j}}})}}

is convergent.

Also, knowing that  y i {\displaystyle y_{i}} tends to zero in the infinity, we can say

∑ j = 1 j = ∞ ( ( − 1 ) j y j ) {\displaystyle \sum _{j=1}^{j=\infty }{((-1)^{j}y_{j})}}

is convergent. The limit of

∑ j = 2 j = i ( ( − 1 ) j x j y j ) = 2 ∑ j = 1 j = i ( ( − 1 ) j + 1 y j ) − y − ( − 1 ) i + 1 y i {\displaystyle \sum _{j=2}^{j=i}{((-1)^{j}{\sqrt {x_{j}y_{j}}})}=2\sum _{j=1}^{j=i}{((-1)^{j+1}y_{j})}-y-(-1)^{i+1}y_{i}}

= 2 ∑ j = 1 j = i ( ( − 1 ) j + 1 x j ) − x − ( − 1 ) i + 1 x i {\displaystyle =2\sum _{j=1}^{j=i}{((-1)^{j+1}x_{j})}-x-(-1)^{i+1}x_{i}}

exists and the series are convergent. It means that for x and y integers and for conditions on the exponents like  n ≥ 3 {\displaystyle n\geq {3}} for Fermat equation :

lim i ⟶ ∞ ( x i ) = x − y = 0 {\displaystyle \lim _{i\longrightarrow {\infty }}{(x_{i})}=x-y=0}

It is confirmed by the fact that the général term of the series tends to zero. Let us prove it. We give two proofs. We must remark that we prove firstly that the following series are convergent, we do not present the proof, here. Let

∑ k = 1 k = 2 m ( ( − 1 ) k + 1 x k e − k 2 m ) {\displaystyle \sum _{k=1}^{k=2m}{((-1)^{k+1}x_{k}e^{-{\frac {k}{\sqrt {2m}}}})}}

= x e − 1 2 m − x 2 e − 2 2 m + x 3 e − 3 2 m − . . . + ( − 1 ) 2 m + 1 x 2 m e − 2 m 2 m {\displaystyle =xe^{-{\frac {1}{\sqrt {2m}}}}-x_{2}e^{-{\frac {2}{\sqrt {2m}}}}+x_{3}e^{-{\frac {3}{\sqrt {2m}}}}-...+(-1)^{2m+1}x_{2m}e^{-{\frac {2m}{\sqrt {2m}}}}}

= x e − 2 2 m + x ( e − 1 2 m − e − 2 2 m ) − x 2 e − 2 2 m + x 3 e − 4 2 m + x 3 ( e − 3 2 m − e − 4 2 m ) − x 4 e − 4 2 m + . . . − x 2 m e − 2 m 2 m {\displaystyle =xe^{-{\frac {2}{\sqrt {2m}}}}+x(e^{-{\frac {1}{\sqrt {2m}}}}-e^{-{\frac {2}{\sqrt {2m}}}})-x_{2}e^{-{\frac {2}{\sqrt {2m}}}}+x_{3}e^{-{\frac {4}{\sqrt {2m}}}}+x_{3}(e^{-{\frac {3}{\sqrt {2m}}}}-e^{-{\frac {4}{\sqrt {2m}}}})-x_{4}e^{-{\frac {4}{\sqrt {2m}}}}+...-x_{2m}e^{-{\frac {2m}{\sqrt {2m}}}}}

= x e − 2 2 m ( e 1 2 m − 1 ) + x 3 e − 4 2 m ( e 1 2 m − 1 ) + . . . + x 2 m − 1 e − 2 m 2 m ( e 1 2 m − 1 ) + {\displaystyle =xe^{-{\frac {2}{\sqrt {2m}}}}(e^{\frac {1}{\sqrt {2m}}}-1)+x_{3}e^{-{\frac {4}{\sqrt {2m}}}}(e^{\frac {1}{\sqrt {2m}}}-1)+...+x_{2m-1}e^{-{\frac {2m}{\sqrt {2m}}}}(e^{\frac {1}{\sqrt {2m}}}-1)+}

+ ( x − x 2 ) e − 2 2 m + ( x 3 − x 4 ) e − 4 2 m + . . . + ( x 2 m − 1 − x 2 m ) e − 2 m 2 m {\displaystyle +(x-x_{2})e^{-{\frac {2}{\sqrt {2m}}}}+(x_{3}-x_{4})e^{-{\frac {4}{\sqrt {2m}}}}+...+(x_{2m-1}-x_{2m})e^{-{\frac {2m}{\sqrt {2m}}}}}

= ( e 1 2 m − 1 ) ( x e − 2 2 m + x 3 e − 4 2 m + . . . + x 2 m − 1 e − 2 m ) + ( x 2 y 2 e − 2 2 m + x 4 y 4 e − 4 2 m + . . . + x 2 m y 2 m e − 2 m 2 m ) {\displaystyle =(e^{\frac {1}{\sqrt {2m}}}-1)(xe^{-{\frac {2}{\sqrt {2m}}}}+x_{3}e^{-{\frac {4}{\sqrt {2m}}}}+...+x_{2m-1}e^{-{\sqrt {2m}}})+({\sqrt {x_{2}y_{2}}}e^{-{\frac {2}{\sqrt {2m}}}}+{\sqrt {x_{4}y_{4}}}e^{-{\frac {4}{\sqrt {2m}}}}+...+{\sqrt {x_{2m}y_{2m}}}e^{-{\frac {2m}{\sqrt {2m}}}})}

= ( e 1 2 m − 1 ) ∑ k = 1 k = m ( x 2 k − 1 e − 2 k 2 m ) + ∑ k = 1 k = m ( x 2 k y 2 k e − 2 k 2 m ) {\displaystyle =(e^{\frac {1}{\sqrt {2m}}}-1)\sum _{k=1}^{k=m}{(x_{2k-1}e^{-{\frac {2k}{\sqrt {2m}}}})}+\sum _{k=1}^{k=m}{({\sqrt {x_{2k}y_{2k}}}e^{-{\frac {2k}{\sqrt {2m}}}})}}

Also

∑ k = 1 k = 2 m ( ( − 1 ) k + 1 y k e − k 2 m ) {\displaystyle \sum _{k=1}^{k=2m}{((-1)^{k+1}y_{k}e^{-{\frac {k}{\sqrt {2m}}}})}}

= y e − 1 2 m − y 2 e − 2 2 m + y 3 e − 3 2 m − . . . + ( − 1 ) 2 m + 1 y 2 m e − 2 m 2 m {\displaystyle =ye^{-{\frac {1}{\sqrt {2m}}}}-y_{2}e^{-{\frac {2}{\sqrt {2m}}}}+y_{3}e^{-{\frac {3}{\sqrt {2m}}}}-...+(-1)^{2m+1}y_{2m}e^{-{\frac {2m}{\sqrt {2m}}}}}

= ( e 1 2 m − 1 ) ∑ k = 1 k = m ( y 2 k − 1 e − 2 k 2 m ) + ∑ k = 1 k = m ( x 2 k y 2 k e − 2 k 2 m ) {\displaystyle =(e^{\frac {1}{\sqrt {2m}}}-1)\sum _{k=1}^{k=m}{(y_{2k-1}e^{-{\frac {2k}{\sqrt {2m}}}})}+\sum _{k=1}^{k=m}{({\sqrt {x_{2k}y_{2k}}}e^{-{\frac {2k}{\sqrt {2m}}}})}}

But

( e 1 2 m − 1 ) ∑ k = 1 k = m ( y 2 k − 1 e − 2 k 2 m ) = S {\displaystyle (e^{\frac {1}{\sqrt {2m}}}-1)\sum _{k=1}^{k=m}{(y_{2k-1}e^{-{\frac {2k}{\sqrt {2m}}}})}=S}

We have

( e 1 2 m − 1 ) ∑ k = 1 k = m ( y 2 k − 1 e − 2 k + 1 2 m ) < S < ( e 1 2 m − 1 ) ∑ k = 1 k = m ( y 2 k − 1 e − 2 k − 1 2 m ) {\displaystyle (e^{\frac {1}{\sqrt {2m}}}-1)\sum _{k=1}^{k=m}{(y_{2k-1}e^{-{\frac {2k+1}{\sqrt {2m}}}})}<S<(e^{\frac {1}{\sqrt {2m}}}-1)\sum _{k=1}^{k=m}{(y_{2k-1}e^{-{\frac {2k-1}{\sqrt {2m}}}})}}

( e 1 2 m − 1 ) e − 3 2 m ∑ k = 1 k = m ( y 2 k − 1 e − 2 k − 2 2 m ) < S < ( e 1 2 m − 1 ) e − 1 2 m ∑ k = 1 k = m ( y 2 k − 1 e − 2 k − 2 2 m ) {\displaystyle (e^{\frac {1}{\sqrt {2m}}}-1)e^{-{\frac {3}{\sqrt {2m}}}}\sum _{k=1}^{k=m}{(y_{2k-1}e^{-{\frac {2k-2}{\sqrt {2m}}}})}<S<(e^{\frac {1}{\sqrt {2m}}}-1)e^{-{\frac {1}{\sqrt {2m}}}}\sum _{k=1}^{k=m}{(y_{2k-1}e^{-{\frac {2k-2}{\sqrt {2m}}}})}}

Thus

lim m ⟶ ∞ ( ( e 1 2 m − 1 ) ∑ k = 1 k = m ( y 2 k − 1 e − 2 k − 2 2 m ) ) = lim m ⟶ ∞ ( ( e 1 2 m − 1 ) y + ( e 1 2 m − 1 ) ∑ k = 2 k = m ( y 2 k − 1 e − 2 k − 2 2 m ) ) {\displaystyle \lim _{m\longrightarrow {\infty }}{((e^{\frac {1}{\sqrt {2m}}}-1)\sum _{k=1}^{k=m}{(y_{2k-1}e^{-{\frac {2k-2}{\sqrt {2m}}}})})}=\lim _{m\longrightarrow {\infty }}{((e^{\frac {1}{\sqrt {2m}}}-1)y+(e^{\frac {1}{\sqrt {2m}}}-1)\sum _{k=2}^{k=m}{(y_{2k-1}e^{-{\frac {2k-2}{\sqrt {2m}}}})})}}

= lim m ⟶ ∞ ( ( e 1 2 m − 1 ) ∑ k = 2 k = m ( y 2 k − 1 e − 2 k − 2 2 m ) ) = lim m ⟶ ∞ ( S ) {\displaystyle =\lim _{m\longrightarrow {\infty }}{((e^{\frac {1}{\sqrt {2m}}}-1)\sum _{k=2}^{k=m}{(y_{2k-1}e^{-{\frac {2k-2}{\sqrt {2m}}}})})}=\lim _{m\longrightarrow {\infty }}{(S)}}

We have

( e 1 2 m − 1 ) ∑ k = p k = m ( y 2 k − 1 e − 2 k − p 2 m ) = A {\displaystyle (e^{\frac {1}{{\sqrt {2}}m}}-1)\sum _{k=p}^{k=m}{(y_{2k-1}e^{-{\frac {2k-p}{\sqrt {2m}}}})}=A}

And

( e 1 2 m − 1 ) ∑ k = p k = m ( y 2 k − 1 e − 2 k − p + 1 2 m ) < A < ( e 1 2 m − 1 ) ∑ k = p k = m ( y 2 k − 1 e − 2 k − p − 1 2 m ) {\displaystyle (e^{\frac {1}{\sqrt {2m}}}-1)\sum _{k=p}^{k=m}{(y_{2k-1}e^{-{\frac {2k-p+1}{\sqrt {2m}}}})}<A<(e^{\frac {1}{\sqrt {2m}}}-1)\sum _{k=p}^{k=m}{(y_{2k-1}e^{-{\frac {2k-p-1}{\sqrt {2m}}}})}}

( e 1 2 m − 1 ) e − 2 2 m ∑ k = p k = m ( y 2 k − 1 e − 2 k − p − 1 2 m ) < A < ( e 1 2 m − 1 ) ∑ k = p k = m ( y 2 k − 1 e − 2 k − p − 1 2 m ) {\displaystyle (e^{\frac {1}{\sqrt {2m}}}-1)e^{-{\frac {2}{\sqrt {2m}}}}\sum _{k=p}^{k=m}{(y_{2k-1}e^{-{\frac {2k-p-1}{\sqrt {2m}}}})}<A<(e^{\frac {1}{\sqrt {2m}}}-1)\sum _{k=p}^{k=m}{(y_{2k-1}e^{-{\frac {2k-p-1}{\sqrt {2m}}}})}}

Hance

lim m ⟶ ∞ ( ( e 1 2 m − 1 ) ∑ k = p k = m ( y 2 k − 1 e − 2 k − p − 1 2 m ) ) {\displaystyle \lim _{m\longrightarrow {\infty }}{((e^{\frac {1}{\sqrt {2m}}}-1)\sum _{k=p}^{k=m}{(y_{2k-1}e^{-{\frac {2k-p-1}{\sqrt {2m}}}})})}}

= lim m ⟶ ∞ ( ( e 1 2 m − 1 ) y 2 p − 1 e − p − 1 2 m + ( e 1 2 m − 1 ) ∑ k = p + 1 k = m ( y 2 k − 1 e − 2 k − p − 1 2 m ) ) {\displaystyle =\lim _{m\longrightarrow {\infty }}{((e^{\frac {1}{\sqrt {2m}}}-1)y_{2p-1}e^{-{\frac {p-1}{\sqrt {2m}}}}+(e^{\frac {1}{\sqrt {2m}}}-1)\sum _{k=p+1}^{k=m}{(y_{2k-1}e^{-{\frac {2k-p-1}{\sqrt {2m}}}})})}}

= lim m ⟶ ∞ ( ( e 1 2 m − 1 ) ∑ k = p + 1 k = m ( y 2 k − 1 e − 2 k − p − 1 2 m ) ) = lim m ⟶ ∞ ( A ) = lim m ⟶ ∞ ( S ) {\displaystyle =\lim _{m\longrightarrow {\infty }}{((e^{\frac {1}{\sqrt {2m}}}-1)\sum _{k=p+1}^{k=m}{(y_{2k-1}e^{-{\frac {2k-p-1}{\sqrt {2m}}}})})}=\lim _{m\longrightarrow {\infty }}{(A)}=\lim _{m\longrightarrow {\infty }}{(S)}}

And

p + 1 = m ⇒ lim m ⟶ ∞ ( A ) = lim m ⟶ ∞ ( S ) = lim m ⟶ ∞ ( ( e 1 2 m − 1 ) y 2 m − 1 e − m 2 m ) = 0 {\displaystyle p+1=m\Rightarrow {\lim _{m\longrightarrow {\infty }}{(A)}=\lim _{m\longrightarrow {\infty }}{(S)}=\lim _{m\longrightarrow {\infty }}{((e^{\frac {1}{\sqrt {2m}}}-1)y_{2m-1}e^{-{\frac {m}{\sqrt {2m}}}})}=0}}

Consequently

lim m ⟶ ∞ ( ( e 1 2 m − 1 ) ∑ k = 1 k = m ( y 2 k − 1 e − 2 k 2 m ) ) = 0 {\displaystyle \lim _{m\longrightarrow {\infty }}{((e^{\frac {1}{\sqrt {2m}}}-1)\sum _{k=1}^{k=m}{(y_{2k-1}e^{-{\frac {2k}{\sqrt {2m}}}})})}=0}

Also

( e 1 2 m − 1 ) ∑ k = 1 k = m ( x 2 k − 1 e − 2 k 2 m ) = S {\displaystyle (e^{\frac {1}{\sqrt {2m}}}-1)\sum _{k=1}^{k=m}{(x_{2k-1}e^{-{\frac {2k}{\sqrt {2m}}}})}=S}

We have

( e 1 2 m − 1 ) ∑ k = 1 k = m ( x 2 k − 1 e − 2 k + 1 2 m ) < S < ( e 1 2 m − 1 ) ∑ k = 1 k = m ( x 2 k − 1 e − 2 k − 1 2 m ) {\displaystyle (e^{\frac {1}{\sqrt {2m}}}-1)\sum _{k=1}^{k=m}{(x_{2k-1}e^{-{\frac {2k+1}{\sqrt {2m}}}})}<S<(e^{\frac {1}{\sqrt {2m}}}-1)\sum _{k=1}^{k=m}{(x_{2k-1}e^{-{\frac {2k-1}{\sqrt {2m}}}})}}

( e 1 2 m − 1 ) e − 3 2 m ∑ k = 1 k = m ( x 2 k − 1 e − 2 k − 2 2 m ) < S < ( e 1 2 m − 1 ) e − 1 2 m ∑ k = 1 k = m ( x 2 k − 1 e − 2 k − 2 2 m ) {\displaystyle (e^{\frac {1}{\sqrt {2m}}}-1)e^{-{\frac {3}{\sqrt {2m}}}}\sum _{k=1}^{k=m}{(x_{2k-1}e^{-{\frac {2k-2}{\sqrt {2m}}}})}<S<(e^{\frac {1}{\sqrt {2m}}}-1)e^{-{\frac {1}{\sqrt {2m}}}}\sum _{k=1}^{k=m}{(x_{2k-1}e^{-{\frac {2k-2}{\sqrt {2m}}}})}}

Thus

lim m ⟶ ∞ ( ( e 1 2 m − 1 ) ∑ k = 1 k = m ( x 2 k − 1 e − 2 k − 2 2 m ) ) = lim m ⟶ ∞ ( ( e 1 2 m − 1 ) x + ( e 1 2 m − 1 ) ∑ k = 2 k = m ( x 2 k − 1 e − 2 k − 2 2 m ) ) {\displaystyle \lim _{m\longrightarrow {\infty }}{((e^{\frac {1}{\sqrt {2m}}}-1)\sum _{k=1}^{k=m}{(x_{2k-1}e^{-{\frac {2k-2}{\sqrt {2m}}}})})}=\lim _{m\longrightarrow {\infty }}{((e^{\frac {1}{\sqrt {2m}}}-1)x+(e^{\frac {1}{\sqrt {2m}}}-1)\sum _{k=2}^{k=m}{(x_{2k-1}e^{-{\frac {2k-2}{\sqrt {2m}}}})})}}

= lim m ⟶ ∞ ( ( e 1 2 m − 1 ) ∑ k = 2 k = m ( x 2 k − 1 e − 2 k − 2 2 m ) ) = lim m ⟶ ∞ ( S ) {\displaystyle =\lim _{m\longrightarrow {\infty }}{((e^{\frac {1}{\sqrt {2m}}}-1)\sum _{k=2}^{k=m}{(x_{2k-1}e^{-{\frac {2k-2}{\sqrt {2m}}}})})}=\lim _{m\longrightarrow {\infty }}{(S)}}

And

( e 1 2 m − 1 ) ∑ k = p k = m ( x 2 k − 1 e − 2 k − p 2 m ) = A {\displaystyle (e^{\frac {1}{{\sqrt {2}}m}}-1)\sum _{k=p}^{k=m}{(x_{2k-1}e^{-{\frac {2k-p}{\sqrt {2m}}}})}=A}

And

( e 1 2 m − 1 ) ∑ k = p k = m ( x 2 k − 1 e − 2 k − p + 1 2 m ) < A < ( e 1 2 m − 1 ) ∑ k = p k = m ( x 2 k − 1 e − 2 k − p − 1 2 m ) {\displaystyle (e^{\frac {1}{\sqrt {2m}}}-1)\sum _{k=p}^{k=m}{(x_{2k-1}e^{-{\frac {2k-p+1}{\sqrt {2m}}}})}<A<(e^{\frac {1}{\sqrt {2m}}}-1)\sum _{k=p}^{k=m}{(x_{2k-1}e^{-{\frac {2k-p-1}{\sqrt {2m}}}})}}

( e 1 2 m − 1 ) e − 2 2 m ∑ k = p k = m ( x 2 k − 1 e − 2 k − p − 1 2 m ) < A < ( e 1 2 m − 1 ) ∑ k = p k = m ( x 2 k − 1 e − 2 k − p − 1 2 m ) {\displaystyle (e^{\frac {1}{\sqrt {2m}}}-1)e^{-{\frac {2}{\sqrt {2m}}}}\sum _{k=p}^{k=m}{(x_{2k-1}e^{-{\frac {2k-p-1}{\sqrt {2m}}}})}<A<(e^{\frac {1}{\sqrt {2m}}}-1)\sum _{k=p}^{k=m}{(x_{2k-1}e^{-{\frac {2k-p-1}{\sqrt {2m}}}})}}

Hence

lim m ⟶ ∞ ( ( e 1 2 m − 1 ) ∑ k = p k = m ( x 2 k − 1 e − 2 k − p − 1 2 m ) ) {\displaystyle \lim _{m\longrightarrow {\infty }}{((e^{\frac {1}{\sqrt {2m}}}-1)\sum _{k=p}^{k=m}{(x_{2k-1}e^{-{\frac {2k-p-1}{\sqrt {2m}}}})})}}

= lim m ⟶ ∞ ( ( e 1 2 m − 1 ) x 2 p − 1 e − p − 1 2 m + ( e 1 2 m − 1 ) ∑ k = p + 1 k = m ( x 2 k − 1 e − 2 k − p − 1 2 m ) ) {\displaystyle =\lim _{m\longrightarrow {\infty }}{((e^{\frac {1}{\sqrt {2m}}}-1)x_{2p-1}e^{-{\frac {p-1}{\sqrt {2m}}}}+(e^{\frac {1}{\sqrt {2m}}}-1)\sum _{k=p+1}^{k=m}{(x_{2k-1}e^{-{\frac {2k-p-1}{\sqrt {2m}}}})})}}

= lim m ⟶ ∞ ( ( e 1 2 m − 1 ) ∑ k = p + 1 k = m ( x 2 k − 1 e − 2 k − p − 1 2 m ) ) = lim m ⟶ ∞ ( A ) = lim m ⟶ ∞ ( S ) {\displaystyle =\lim _{m\longrightarrow {\infty }}{((e^{\frac {1}{\sqrt {2m}}}-1)\sum _{k=p+1}^{k=m}{(x_{2k-1}e^{-{\frac {2k-p-1}{\sqrt {2m}}}})})}=\lim _{m\longrightarrow {\infty }}{(A)}=\lim _{m\longrightarrow {\infty }}{(S)}}

p + 1 = m ⇒ lim m ⟶ ∞ ( A ) = lim m ⟶ ∞ ( S ) = lim m ⟶ ∞ ( ( e 1 2 m − 1 ) x 2 m − 1 e − m 2 m ) = 0 {\displaystyle p+1=m\Rightarrow {\lim _{m\longrightarrow {\infty }}{(A)}=\lim _{m\longrightarrow {\infty }}{(S)}=\lim _{m\longrightarrow {\infty }}{((e^{\frac {1}{\sqrt {2m}}}-1)x_{2m-1}e^{-{\frac {m}{\sqrt {2m}}}})}=0}}

Consequently

lim m ⟶ ∞ ( ( e 1 2 m − 1 ) ∑ k = 1 k = m ( x 2 k − 1 e − 2 k 2 m ) ) = 0 {\displaystyle \lim _{m\longrightarrow {\infty }}{((e^{\frac {1}{\sqrt {2m}}}-1)\sum _{k=1}^{k=m}{(x_{2k-1}e^{-{\frac {2k}{\sqrt {2m}}}})})}=0}

We deduce

0 < lim m ⟶ ∞ ( ∑ k = 1 k = 2 m ( ( − 1 ) k + 1 x k e − k 2 m ) ) {\displaystyle 0<\lim _{m\longrightarrow {\infty }}{(\sum _{k=1}^{k=2m}{((-1)^{k+1}x_{k}e^{-{\frac {k}{\sqrt {2m}}}})})}}

= lim m ⟶ ∞ ( ∑ k = 1 k = 2 m ( ( − 1 ) k + 1 x k e − k 2 m ) ) {\displaystyle =\lim _{m\longrightarrow {\infty }}{(\sum _{k=1}^{k=2m}{((-1)^{k+1}x_{k}e^{-{\frac {k}{\sqrt {2m}}}})})}}

= lim m ⟶ ∞ ( ∑ k = 1 k = m ( x 2 k y 2 k e − 2 k 2 m ) ) < lim m ⟶ ∞ ( ∑ k = 1 k = m ( x 2 k y 2 k ) ) {\displaystyle =\lim _{m\longrightarrow {\infty }}{(\sum _{k=1}^{k=m}{({\sqrt {x_{2k}y_{2k}}}e^{-{\frac {2k}{\sqrt {2m}}}})})}<\lim _{m\longrightarrow {\infty }}{(\sum _{k=1}^{k=m}{({\sqrt {x_{2k}y_{2k}}})})}}

< lim m ⟶ ∞ ( ∑ k = 1 k = 2 m ( x k y k ) ) = y {\displaystyle <\lim _{m\longrightarrow {\infty }}{(\sum _{k=1}^{k=2m}{({\sqrt {x_{k}y_{k}}})})}=y}

Thus

lim m ⟶ ∞ ( ∑ k = 1 k = 2 m ( ( − 1 ) k + 1 ( x k − y k ) e − k 2 m ) ) = 0 {\displaystyle \lim _{m\longrightarrow {\infty }}{(\sum _{k=1}^{k=2m}{((-1)^{k+1}(x_{k}-y_{k})e^{-{\frac {k}{\sqrt {2m}}}})})}=0}

= lim m ⟶ ∞ ( ( x − y ) ∑ k = 1 k = 2 m ( e − k 2 m ) ) = lim m ⟶ ∞ ( ( x − y ) e − 1 2 m 1 − e − 2 m 1 + e − 1 2 m ) = x − y 2 = 0 {\displaystyle =\lim _{m\longrightarrow {\infty }}{((x-y)\sum _{k=1}^{k=2m}{(e^{-{\frac {k}{\sqrt {2m}}}})})}=\lim _{m\longrightarrow {\infty }}{((x-y)e^{-{\frac {1}{\sqrt {2m}}}}{\frac {1-e^{-{\sqrt {2m}}}}{1+e^{-{\frac {1}{\sqrt {2m}}}}}})}={\frac {x-y}{2}}=0}

And

x − y = 0 {\displaystyle x-y=0}

the only solution is  x y ( x − y ) = 0 {\displaystyle xy(x-y)=0} ,  x y = 0 {\displaystyle xy=0} if at least one of the sequences  x i {\displaystyle x_{i}} or  y i {\displaystyle y_{i}} is constant. And the second proof. Let

x i ( t ) = ∑ k = 2 k = i − 1 ( x k e − k t ) = ∑ k = 2 k = i ( x k e − k t ) − x i e − i t = x ′ i ( t ) − x i e − i t {\displaystyle x_{i}(t)=\sum _{k=2}^{k=i-1}{(x_{k}e^{-kt})}=\sum _{k=2}^{k=i}{(x_{k}e^{-kt})}-x_{i}e^{-it}={x'}_{i}(t)-x_{i}e^{-it}}

y i ( t ) = ∑ k = 2 k = i − 1 ( y k e − k t ) = ∑ k = 2 k = i ( y k e − k t ) − y i e − i t = y ′ i ( t ) − y i e − i t {\displaystyle y_{i}(t)=\sum _{k=2}^{k=i-1}{(y_{k}e^{-kt})}=\sum _{k=2}^{k=i}{(y_{k}e^{-kt})}-y_{i}e^{-it}={y'}_{i}(t)-y_{i}e^{-it}}

u i ( t ) = ∑ k = 2 k = i ( x k y k e − k t ) = ∑ k = 2 k = i − 1 ( ( x k − 1 − x k ) e − k t ) {\displaystyle u_{i}(t)=\sum _{k=2}^{k=i}{({\sqrt {x_{k}y_{k}}}e^{-kt})}=\sum _{k=2}^{k=i-1}{((x_{k-1}-x_{k})e^{-kt})}}

= x e − 2 t + x 2 ( − e − 2 t + e − 3 t ) + x 3 ( − e − 3 t + e − 4 t ) + . . . + x i − 1 ( − e − ( i − 1 ) t + e − i t ) − x i e − i t {\displaystyle =xe^{-2t}+x_{2}(-e^{-2t}+e^{-3t})+x_{3}(-e^{-3t}+e^{-4t})+...+x_{i-1}(-e^{-(i-1)t}+e^{-it})-x_{i}e^{-it}}

= x e − 2 t + x 2 e − 2 t ( − 1 + e − t ) + x 3 e − 3 t ( − 1 + e − t ) + . . . + x i − 1 e − ( i − 1 ) t ( − 1 + e − t ) − x i e − i t {\displaystyle =xe^{-2t}+x_{2}e^{-2t}(-1+e^{-t})+x_{3}e^{-3t}(-1+e^{-t})+...+x_{i-1}e^{-(i-1)t}(-1+e^{-t})-x_{i}e^{-it}}

= x e − 2 t − x i e − i t + ( − 1 + e − t ) ( x 2 e − 2 t + x 3 e − 3 t + . . . + x i e − ( i − 1 ) t ) {\displaystyle =xe^{-2t}-x_{i}e^{-it}+(-1+e^{-t})(x_{2}e^{-2t}+x_{3}e^{-3t}+...+x_{i}e^{-(i-1)t})}

= x e − 2 t − x i e − i t + ( − 1 + e − t ) x i ( t ) = x e − 2 t − x i e − i t + ( − 1 + e − t ) x ′ i ( t ) − ( − 1 + e − t ) x i e − i t = x e − 2 t − x i e − ( i + 1 ) t + ( − 1 + e − t ) x ′ i ( t ) {\displaystyle =xe^{-2t}-x_{i}e^{-it}+(-1+e^{-t})x_{i}(t)=xe^{-2t}-x_{i}e^{-it}+(-1+e^{-t}){x'}_{i}(t)-(-1+e^{-t})x_{i}e^{-it}=xe^{-2t}-x_{i}e^{-(i+1)t}+(-1+e^{-t}){x'}_{i}(t)}

let

t = 1 i ⇒ lim i ⟶ ∞ ( u i ( 1 i ) ) = lim i ⟶ ∞ ( ∑ k = 2 k = i ( x k y k e − k i ) ) = lim i ⟶ ∞ ( ∑ k = 2 k = i ( x k y k ) ) = y {\displaystyle t={\frac {1}{\sqrt {i}}}\Rightarrow {\lim _{i\longrightarrow {\infty }}{(u_{i}({\frac {1}{\sqrt {i}}}))}=\lim _{i\longrightarrow {\infty }}{(\sum _{k=2}^{k=i}{({\sqrt {x_{k}y_{k}}}e^{-{\frac {k}{\sqrt {i}}}})})}}=\lim _{i\longrightarrow {\infty }}{(\sum _{k=2}^{k=i}{({\sqrt {x_{k}y_{k}}})})}=y}

= x + lim i ⟶ ∞ ( ( − 1 + e − 1 i ) ∑ k = 2 k = i − 1 ( x k e − k i ) − x i e − i ) = x {\displaystyle =x+\lim _{i\longrightarrow {\infty }}{((-1+e^{-{\frac {1}{\sqrt {i}}}})\sum _{k=2}^{k=i-1}{(x_{k}e^{-{\frac {k}{\sqrt {i}}}})}-x_{i}e^{-{\sqrt {i}}})}=x}

= y + lim i ⟶ ∞ ( ( − 1 + e − 1 i ) ∑ k = 2 k = i − 1 ( y k e − k i ) − y i e − i ) = y {\displaystyle =y+\lim _{i\longrightarrow {\infty }}{((-1+e^{-{\frac {1}{\sqrt {i}}}})\sum _{k=2}^{k=i-1}{(y_{k}e^{-{\frac {k}{\sqrt {i}}}})}-y_{i}e^{-{\sqrt {i}}})}=y}

We Recapitulate

x i > x i + 1 , y i > y i + 1 ⇒ x = y {\displaystyle x_{i}>x_{i+1},\quad {y_{i}>y_{i+1}}\Rightarrow {x=y}}

x i = x i + 1 = x i + 1 + x i + 1 y i + 1 ⇒ x y = 0 {\displaystyle x_{i}=x_{i+1}=x_{i+1}+{\sqrt {x_{i+1}y_{i+1}}}\Rightarrow {xy=0}}

y i = y i + 1 = y i + 1 + x i + 1 y i + 1 ⇒ x y = 0 {\displaystyle y_{i}=y_{i+1}=y_{i+1}+{\sqrt {x_{i+1}y_{i+1}}}\Rightarrow {xy=0}}

⇒ x y ( x − y ) = 0 {\displaystyle \Rightarrow {xy(x-y)=0}}

is the only solution of (1) and (2). This result is paradoxal, we remark that we have not put any condition on n, because there are solutions for  n ≤ 2 {\displaystyle n\leq {2}} . The answer is related to Matiasevic theorem which claims that dos not exist an algorithm to prove theorems related to diophantine equations and we gave one : The approach must conduct then to an impossibility. We confirm Matiasevic theorem and prove it because our algorithm is available for n=1. The approach is more important than it appears, it is an answer to problems more general than Fermat theorem or Beal or Fermat-Catalan conjectures. We will try to prove them. The series become

∑ k = 2 k = ∞ ( ( − 1 ) k x k y k ) = ∑ k = 2 k = ∞ ( ( − 1 ) k x k ) = ∑ k = 2 k = ∞ ( ( − 1 ) k y k ) {\displaystyle \sum _{k=2}^{k={\infty }}{((-1)^{k}{\sqrt {x_{k}y_{k}}})}=\sum _{k=2}^{k={\infty }}{((-1)^{k}x_{k})}=\sum _{k=2}^{k={\infty }}{((-1)^{k}y_{k})}}

= x 2 − x 3 + x 4 − x 5 + . . . {\displaystyle =x_{2}-x_{3}+x_{4}-x_{5}+...}

= 2 ∑ k = 1 k = ∞ ( ( − 1 ) k + 1 x k ) − x = x − 2 x 2 + 2 x 3 − 2 x 4 + . . . {\displaystyle =2\sum _{k=1}^{k={\infty }}{((-1)^{k+1}x_{k})}-x=x-2x_{2}+2x_{3}-2x_{4}+...}

= 2 ∑ k = 1 k = ∞ ( ( − 1 ) k + 1 y k ) − y = y − 2 y 2 + 2 y 3 − 2 y 4 + . . . {\displaystyle =2\sum _{k=1}^{k={\infty }}{((-1)^{k+1}y_{k})}-y=y-2y_{2}+2y_{3}-2y_{4}+...}

⇒ 3 ( x 2 − x 3 + x 4 − x 5 + . . . ) = 3 ∑ k = 2 k = ∞ ( ( − 1 ) k x k ) = 3 ∑ k = 2 k = ∞ ( ( − 1 ) k y k ) = x = y {\displaystyle \Rightarrow {3(x_{2}-x_{3}+x_{4}-x_{5}+...)=3\sum _{k=2}^{k={\infty }}{((-1)^{k}x_{k})}=3\sum _{k=2}^{k={\infty }}{((-1)^{k}y_{k})}=x=y}}

⇒ x 2 − x 3 + x 4 − x 5 + . . . = ∑ k = 2 k = ∞ ( ( − 1 ) k x k ) = ∑ k = 2 k = ∞ ( ( − 1 ) k y k ) = x 3 = y 3 {\displaystyle \Rightarrow {x_{2}-x_{3}+x_{4}-x_{5}+...=\sum _{k=2}^{k={\infty }}{((-1)^{k}x_{k})}=\sum _{k=2}^{k={\infty }}{((-1)^{k}y_{k})}={\frac {x}{3}}={\frac {y}{3}}}}

and

∑ k = 2 k = ∞ ( x k y k ) = ∑ k = 2 k = ∞ ( x k ) = ∑ k = 2 k = i ( y k ) {\displaystyle \sum _{k=2}^{k={\infty }}{({\sqrt {x_{k}y_{k}}})}=\sum _{k=2}^{k={\infty }}{(x_{k})}=\sum _{k=2}^{k={i}}{(y_{k})}}

= x 2 + x 3 + x 4 + x 5 + . . . = y 2 + y 3 + y 4 + y 5 + . . . = y = x {\displaystyle =x_{2}+x_{3}+x_{4}+x_{5}+...=y_{2}+y_{3}+y_{4}+y_{5}+...=y=x}

and

x i = x 2 i − 1 ∏ j = 0 j = i − 2 ( x 2 j + y 2 j ) − 1 = y i {\displaystyle x_{i}=x^{2^{i-1}}\prod _{j=0}^{j=i-2}{(x^{2^{j}}+y^{2^{j}})^{-1}}=y_{i}}

= x 2 i − 1 ∏ j = 0 j = i − 2 ( 2 x 2 j ) − 1 = x 2 i − 1 2 i − 1 x 2 i − 1 − 1 {\displaystyle =x^{2^{i-1}}\prod _{j=0}^{j=i-2}{(2x^{2^{j}})^{-1}}={\frac {x^{2^{i-1}}}{2^{i-1}x^{2^{i-1}-1}}}}

= x i = x 2 i − 1 = y i = y 2 i − 1 {\displaystyle =x_{i}={\frac {x}{2^{i-1}}}=y_{i}={\frac {y}{2^{i-1}}}}

This development is in fact a test of impossibility. The sequences and series are a consequence of Fermat equation and of other diophantine equations (as we will see). The question now is : why are there solutions for n=2 ? The answer is in the formulas, as seen. It is important to note that for n=1, there are trivial solutions. But, for n=1, lemma 3 allows to write

x i = x 2 i − 1 x 2 i − 1 − y 2 i − 1 ( x − y ) = U X 2 i − 1 X 2 i − 1 − Y 2 i − 1 ( X − Y ) {\displaystyle x_{i}={\frac {x^{2^{i-1}}}{x^{2^{i-1}}-y^{2^{i-1}}}}(x-y)=U{\frac {X^{2^{i-1}}}{X^{2^{i-1}}-Y^{2^{i-1}}}}(X-Y)}

= U X 2 2 i − 2 X 2 2 i − 2 − Y 2 2 i − 2 ( X − Y ) {\displaystyle =U{\frac {X^{2{2^{i-2}}}}{X^{2{2^{i-2}}}-Y^{2{2^{i-2}}}}}(X-Y)}

and

y i = y 2 i − 1 x 2 i − 1 − y 2 i − 1 ( x − y ) = U Y 2 i − 1 X 2 i − 1 − Y 2 i − 1 ( X − Y ) {\displaystyle y_{i}={\frac {y^{2^{i-1}}}{x^{2^{i-1}}-y^{2^{i-1}}}}(x-y)=U{\frac {Y^{2^{i-1}}}{X^{2^{i-1}}-Y^{2^{i-1}}}}(X-Y)}

= U Y 2 2 i − 2 X 2 2 i − 2 − Y 2 2 i − 2 ( X − Y ) {\displaystyle =U{\frac {Y^{2{2^{i-2}}}}{X^{2{2^{i-2}}}-Y^{2{2^{i-2}}}}}(X-Y)}

and

u i = x i + y i = U X 2 2 i − 2 + Y 2 2 i − 2 X 2 2 i − 2 − Y 2 2 i − 2 ( X − Y ) = X 2 2 i − 2 + Y 2 2 i − 2 X 2 2 i − 2 − Y 2 2 i − 2 ( X 2 − Y 2 ) {\displaystyle u_{i}=x_{i}+y_{i}=U{\frac {X^{2{2^{i-2}}}+Y^{2{2^{i-2}}}}{X^{2{2^{i-2}}}-Y^{2{2^{i-2}}}}}(X-Y)={\frac {X^{2{2^{i-2}}}+Y^{2{2^{i-2}}}}{X^{2{2^{i-2}}}-Y^{2{2^{i-2}}}}}(X^{2}-Y^{2})}

It is the expression of  u ′ i − 1 {\displaystyle {u'}_{i-1}} , the  u i − 1 {\displaystyle u_{i-1}} exponent 2.

The case n=1 conducts to the case n=2 and as there are solutions for n=1, it will be the same for n=2 !

For n=4

u i = x i + y i = X 4 2 i − 3 + Y 4 2 i − 3 X 4 2 i − 3 − Y 4 2 i − 3 U ( X − Y ) {\displaystyle u_{i}=x_{i}+y_{i}={\frac {X^{4{2^{i-3}}}+Y^{4{2^{i-3}}}}{X^{4{2^{i-3}}}-Y^{4{2^{i-3}}}}}U(X-Y)}

the case n=4 is different, in this formula the exponent i-3 does not guarantee the existence of the sequences if

i=2.

Then, the case n=2 is the only exception. The only solution for n>2 is xy(x-y)=0, there is no solution.

Another application is Beal equation. It is

U c = X a + Y b {\displaystyle U^{c}=X^{a}+Y^{b}}

G C D ( X , Y ) = 1 {\displaystyle GCD(X,Y)=1}

We pose

u = U 2 c {\displaystyle u=U^{2c}}

x = U c X a {\displaystyle x=U^{c}X^{a}}

y = U c Y b {\displaystyle y=U^{c}Y^{b}}

z = X a Y b {\displaystyle z=X^{a}Y^{b}}

and

u = U 2 c = U c ( X a + Y b ) = x + y {\displaystyle u=U^{2c}=U^{c}(X^{a}+Y^{b})=x+y}

and

1 z = 1 X a Y b = U 2 c U c X a U c Y b = u x y = x + y x y = 1 x + 1 y {\displaystyle {\frac {1}{z}}={\frac {1}{X^{a}Y^{b}}}={\frac {U^{2c}}{U^{c}X^{a}U^{c}Y^{b}}}={\frac {u}{xy}}={\frac {x+y}{xy}}={\frac {1}{x}}+{\frac {1}{y}}}

it is lemma 1 and, with

U c = X a + Y b {\displaystyle U^{c}=X^{a}+Y^{b}}

the solutions are

x y ( x − y ) = U 2 c X a Y b ( U c X a − U c Y b ) = 0 {\displaystyle xy(x-y)=U^{2c}X^{a}Y^{b}(U^{c}X^{a}-U^{c}Y^{b})=0}

or impossible solutions for a>2 et b>2 et c>2\.

Another application is the following equation.

U n = X 1 n 1 + X 2 n 2 + . . . + X i n i {\displaystyle U^{n}=X_{1}^{n_{1}}+X_{2}^{n_{2}}+...+X_{i}^{n_{i}}}

G C D ( X j , X k ) = 1 ; ∀ j , k , j ≠ k {\displaystyle GCD(X_{j},X_{k})=1;\forall {j,k,j\neq {k}}}

We conjecture and prove that there is no solutions for n>i(i-1) and  n j > i ( i − 1 ) {\displaystyle n_{j}>i(i-1)} ,  ∀ j ∈ { 1 , 2 , . . . , i } {\displaystyle \forall {j\in \\{1,2,...,i\\}}} . We can not know when there are solutions as proved by Matiasevic when one of the exponent is less or equal to i(i-1).

## LEMMA 6

[edit | edit source]

The solution  ∀ k ≥ 2 {\displaystyle \forall {k\geq {2}}} is

X k = 0 {\displaystyle X_{k}=0}

## Proof of lemma 6

[edit | edit source]

We pose

u = U 2 n {\displaystyle u=U^{2n}}

x = U n X k n k {\displaystyle x=U^{n}X_{k}^{n_{k}}}

y = U n ( U n − X k n k ) {\displaystyle y=U^{n}(U^{n}-X_{k}^{n_{k}})}

z = X k n k ( U n − X k n k ) {\displaystyle z=X_{k}^{n_{k}}(U^{n}-X_{k}^{n_{k}})}

u = x + y {\displaystyle u=x+y}

and

1 z = 1 x + 1 y {\displaystyle {\frac {1}{z}}={\frac {1}{x}}+{\frac {1}{y}}}

it is lemma 1. Its solution is

U = X k = 0 ; ∀ k ∈ { 1 , 2 , . . . , i } {\displaystyle U=X_{k}=0;\forall {k\in {\\{1,2,...,i\\}}}}

But, why are not they solutions for n>i(i-1) and  n j > i ( i − 1 ) {\displaystyle n_{j}>i(i-1)}  ?

We will generalize the definition of the sequences.

We will define general sequences.

Our goal is to prove that if  X i {\displaystyle X_{i}} ,

( i ≥ 2 {\displaystyle i\geq {2}} ),  n i {\displaystyle n_{i}} ,  U {\displaystyle U} ,  n {\displaystyle n} ,  P G C D ( X 1 , X 2 , . . . , X i ) = 1 {\displaystyle PGCD(X_{1},X_{2},...,X_{i})=1}

are positive integers, then

X a = X b = 0 ; ∀ a , b = 1 , 2 , . . . , i {\displaystyle X_{a}=X_{b}=0;\forall {a,b=1,2,...,i}}

n k > i ( i − 1 ) , ∀ k = 1 , 2 , . . . , i , n > i ( i − 1 ) {\displaystyle n_{k}>{i(i-1)},\forall {k=1,2,...,i},n>i(i-1)}

for the equation

X 1 n 1 + X 2 n 2 + . . . + X i n i = U n ( e ) {\displaystyle X_{1}^{n_{1}}+X_{2}^{n_{2}}+...+X_{i}^{n_{i}}=U^{n}\quad \quad {(e)}}

When  n ≤ i ( i − 1 ) , n k ≤ i ( i − 1 ) , k = 1 , 2 , . . . , i {\displaystyle n\leq {i(i-1)},n_{k}\leq {i(i-1)},k=1,2,...,i}

there are solutions, for example :

i = 2 {\displaystyle i=2} has  3 2 + 4 2 = 5 2 {\displaystyle 3^{2}+4^{2}=5^{2}}

and  i = 3 {\displaystyle i=3} has  3 3 + 4 3 + 5 3 = 6 3 {\displaystyle 3^{3}+4^{3}+5^{3}=6^{3}}

and

95800 4 + 217519 4 + 414560 4 = 422481 4 {\displaystyle 95800^{4}+217519^{4}+414560^{4}=422481^{4}}

and  i = 4 {\displaystyle i=4} has  27 5 + 84 5 + 110 5 + 133 5 = 144 5 {\displaystyle 27^{5}+84^{5}+110^{5}+133^{5}=144^{5}}

etc...

We suppose (e) verified and that  G C D ( X k ) = 1 {\displaystyle GCD(X_{k})=1}

k ∈ { 1 , 2 , . . . , i } ; i ≥ 2 {\displaystyle k\in \\{1,2,...,i\\};i\geq {2}} , soit

x k = U ( i − 1 ) n X k n k {\displaystyle x_{k}=U^{(i-1)n}X_{k}^{n_{k}}}

k ∈ { 1 , 2 , . . . , i } {\displaystyle k\in {\\{1,2,...,i\\}}}

and

u = U i n {\displaystyle u=U^{in}}

and

v = X 1 n 1 X 2 n 2 . . . X i n i {\displaystyle v=X_{1}^{n_{1}}X_{2}^{n_{2}}...X_{i}^{n_{i}}}

## LEMMA 7

[edit | edit source]

x k {\displaystyle x_{k}} ,  k = 1 , 2 , . . . i {\displaystyle k=1,2,...i} , u, v verify

x 1 + x 2 + . . . + x i = U ( i − 1 ) n ( X 1 n 1 + X 2 n 2 + . . . + X i n i ) = U i n = u ( 3 ) {\displaystyle x_{1}+x_{2}+...+x_{i}=U^{(i-1)n}(X_{1}^{n_{1}}+X_{2}^{n_{2}}+...+X_{i}^{n_{i}})=U^{in}=u\quad \quad {(3)}}

and

1 v = 1 X 1 n 1 X 2 n 2 . . . X i n i = U i ( i − 1 ) n U ( i − 1 ) n X 1 n 1 U ( i − 1 ) n X 2 n 2 . . . U ( i − 1 ) n X i n i = u ( i − 1 ) x 1 x 2 . . . x i ( 4 ) {\displaystyle {\frac {1}{v}}={\frac {1}{X_{1}^{n_{1}}X_{2}^{n_{2}}...X_{i}^{n_{i}}}}={\frac {U^{{i(i-1)}n}}{U^{(i-1)n}X_{1}^{n_{1}}U^{(i-1)n}X_{2}^{n_{2}}...U^{(i-1)n}X_{i}^{n_{i}}}}={\frac {u^{(i-1)}}{x_{1}x_{2}...x_{i}}}\quad \quad {(4)}}

We pose

x k , 0 = x k {\displaystyle x_{k,0}=x_{k}}

u 0 = u {\displaystyle u_{0}=u}

v 0 = v {\displaystyle v_{0}=v}

and

x k , 1 = x k i ( x 1 + x 2 + . . . + x i ) − ( i − 1 ) {\displaystyle x_{k,1}=x_{k}^{i}(x_{1}+x_{2}+...+x_{i})^{-(i-1)}}

k = 1 , 2 , . . . , i {\displaystyle k=1,2,...,i}

it implies

u = x 1 + x 2 + . . . + x i = ( x 1 , 1 1 i + x 2 , 1 1 i + . . . + x i , 1 1 i ) i > u 1 > 1 {\displaystyle u=x_{1}+x_{2}+...+x_{i}=(x_{1,1}^{\frac {1}{i}}+x_{2,1}^{\frac {1}{i}}+...+x_{i,1}^{\frac {1}{i}})^{i}>u_{1}>1}

and

x k , 0 = x k = x k , 1 1 i ( x 1 + x 2 + . . . + x i ) ( i − 1 ) i {\displaystyle x_{k,0}=x_{k}=x_{k,1}^{\frac {1}{i}}(x_{1}+x_{2}+...+x_{i})^{\frac {(i-1)}{i}}}

= x k , 1 1 i ( x 1 , 1 1 i + x 2 , 1 1 i + . . . + x i , 1 1 i ) ( i − 1 ) > x k , 1 > 0 {\displaystyle =x_{k,1}^{\frac {1}{i}}(x_{1,1}^{\frac {1}{i}}+x_{2,1}^{\frac {1}{i}}+...+x_{i,1}^{\frac {1}{i}})^{(i-1)}>x_{k,1}>0}

and

v = x 1 , 0 x 2 , 0 . . . x i , 0 u ( i − 1 ) = x 1 , 1 1 i x 2 , 1 1 i . . . x i , 1 1 i > v 1 = x 1 , 1 x 2 , 1 . . . x i , 1 u 1 ( i − 1 ) > 0 {\displaystyle v={\frac {x_{1,0}x_{2,0}...x_{i,0}}{u^{(i-1)}}}=x_{1,1}^{\frac {1}{i}}x_{2,1}^{\frac {1}{i}}...x_{i,1}^{\frac {1}{i}}>v_{1}={\frac {x_{1,1}x_{2,1}...x_{i,1}}{u_{1}^{(i-1)}}}>0}

The reasoning is available until infinity. Then

u j = x 1 , j + x 2 , j + . . . + x i , j = ( x 1 , j + 1 1 i + x 2 , j + 1 1 i + . . . + x i , j + 1 1 i ) i > u j + 1 > 0 {\displaystyle u_{j}=x_{1,j}+x_{2,j}+...+x_{i,j}=(x_{1,{j+1}}^{\frac {1}{i}}+x_{2,{j+1}}^{\frac {1}{i}}+...+x_{i,{j+1}}^{\frac {1}{i}})^{i}>u_{j+1}>0}

and

x k , j = x k , j + 1 1 i ( x 1 , j + 1 1 i + x 2 , j + 1 1 i + . . . + x i , j + 1 1 i ) ( i − 1 ) > x k , j + 1 > 0 {\displaystyle x_{k,j}=x_{k,{j+1}}^{\frac {1}{i}}(x_{1,{j+1}}^{\frac {1}{i}}+x_{2,{j+1}}^{\frac {1}{i}}+...+x_{i,{j+1}}^{\frac {1}{i}})^{(i-1)}>x_{k,{j+1}}>0}

and

k = 1 , 2 , . . . , i {\displaystyle k=1,2,...,i}

and

v j = x 1 , j x 2 , j . . . x i , j u j ( i − 1 ) = x 1 , j + 1 1 i x 2 , j + 1 1 i . . . x i , j + 1 1 i > v j + 1 = x 1 , j + 1 x 2 , j + 1 . . . x i , j + 1 u j + 1 ( i − 1 ) > 0 {\displaystyle v_{j}={\frac {x_{1,j}x_{2,j}...x_{i,j}}{u_{j}^{(i-1)}}}=x_{1,{j+1}}^{\frac {1}{i}}x_{2,{j+1}}^{\frac {1}{i}}...x_{i,{j+1}}^{\frac {1}{i}}>v_{j+1}={\frac {x_{1,{j+1}}x_{2,{j+1}}...x_{i,{j+1}}}{u_{j+1}^{(i-1)}}}>0}

x k , j , v j , u j {\displaystyle x_{k,j},v_{j},u_{j}} are positive  ∀ j > 1 {\displaystyle \forall {j>1}} ,  ∀ k = 1 , 2 , . . . , i {\displaystyle \forall {k=1,2,...,i}} .

## LEMMA 8

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(P) is the expression :

x k , j = x k i j ( ∏ l = 0 l = j − 1 x 1 i l + x 2 i l + . . . + x i i l ) − ( i − 1 ) {\displaystyle x_{k,j}=x_{k}^{i^{j}}(\prod _{l=0}^{l={j-1}}{x_{1}^{i^{l}}+x_{2}^{i^{l}}+...+x_{i}^{i^{l}}})^{-(i-1)}}

## Proof of lemma 8

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for j=1, it is verified because of the definition of  x k , 1 , u 1 , v 1 {\displaystyle x_{k,1},u_{1},v_{1}} ,

we suppose (P) true and the expression of  u j {\displaystyle u_{j}} implies, with (P), that

x k , j = x k , j + 1 1 i ( x 1 , j + 1 1 i + x 2 , j + 1 1 3 + . . . + x i , j + 1 1 i ) ( i − 1 ) {\displaystyle x_{k,j}=x_{k,{j+1}}^{\frac {1}{i}}(x_{1,{j+1}}^{\frac {1}{i}}+x_{2,{j+1}}^{\frac {1}{3}}+...+x_{i,{j+1}}^{\frac {1}{i}})^{(i-1)}}

x k , j + 1 1 i = x k , j ( x 1 , j + 1 1 i + x 2 , j + 1 1 i + . . . + x i , j + 1 1 i ) − ( i − 1 ) {\displaystyle x_{k,{j+1}}^{\frac {1}{i}}=x_{k,j}(x_{1,{j+1}}^{\frac {1}{i}}+x_{2,{j+1}}^{\frac {1}{i}}+...+x_{i,{j+1}}^{\frac {1}{i}})^{-(i-1)}}

and

x k , j + 1 = x k , j i ( x 1 , j + 1 i + x 2 , j + 1 1 i + . . . + x i , j + 1 1 i ) − ( i − 1 ) i = x k , j i ( x 1 , j + x 2 , j + . . . + x i , j ) − ( i − 1 ) {\displaystyle x_{k,{j+1}}=x_{k,j}^{i}(x_{1,{j+1}}^{\frac {}{i}}+x_{2,{j+1}}^{\frac {1}{i}}+...+x_{i,{j+1}}^{\frac {1}{i}})^{-(i-1)i}=x_{k,j}^{i}(x_{1,j}+x_{2,j}+...+x_{i,j})^{-(i-1)}}

= x k i j + 1 ∏ l = 0 l = j − 1 ( x 1 i l + x 2 i l + . . . + x i i l ) − i ( i − 1 ) ( x 1 i j + x 2 i j + . . . + x i i j ) − ( i − 1 ) ∏ l = 0 l = j − 1 ( x 1 i l + x 2 i l + . . . + x i i l ) ( i − 1 ) 2 {\displaystyle =x_{k}^{i^{j+1}}\prod _{l=0}^{l={j-1}}{(x_{1}^{i^{l}}+x_{2}^{i^{l}}+...+x_{i}^{i^{l}})^{-{i(i-1)}}(x_{1}^{i^{j}}+x_{2}^{i^{j}}+...+x_{i}^{i^{j}})^{-(i-1)}}\prod _{l=0}^{l={j-1}}{(x_{1}^{i^{l}}+x_{2}^{i^{l}}+...+x_{i}^{i^{l}})^{(i-1)^{2}}}}

= x k i j + 1 ∏ l = 0 l = j ( x 1 i l + x 2 i l + . . . + x i i l ) − ( i − 1 ) {\displaystyle =x_{k}^{i^{j+1}}\prod _{l=0}^{l=j}{(x_{1}^{i^{l}}+x_{2}^{i^{l}}+...+x_{i}^{i^{l}})^{-(i-1)}}}

then

x k , j = x k i j ∏ l = 0 l = j − 1 ( x 1 i l + x 2 i l + . . . + x i i l ) − ( i − 1 ) {\displaystyle x_{k,j}=x_{k}^{i^{j}}\prod _{l=0}^{l={j-1}}{(x_{1}^{i^{l}}+x_{2}^{i^{l}}+...+x_{i}^{i^{l}})^{-(i-1)}}}

Why are not they solutions for  n > i ( i − 1 ) , n k > i ( i − 1 ) {\displaystyle n>i(i-1),n_{k}>i(i-1)}  ?

We suppose

n = n k = i ( i − 1 ) {\displaystyle n=n_{k}=i(i-1)}

the formula becomes

x k , j = x k i j ( ∏ l = 0 l = j − 1 x 1 i l + x 2 i l + . . . + x i i l ) − ( i − 1 ) {\displaystyle x_{k,j}=x_{k}^{i^{j}}(\prod _{l=0}^{l={j-1}}{x_{1}^{i^{l}}+x_{2}^{i^{l}}+...+x_{i}^{i^{l}}})^{-(i-1)}}

= U n i j X k n k i j ( ∏ l = 0 l = j − 1 ( U n i l X 1 n 1 i l + U n i l X 2 n 2 i l + . . . + U n i l X i n i i l ) ) − ( i − 1 ) {\displaystyle =U^{n{i^{j}}}X_{k}^{n_{k}{i^{j}}}(\prod _{l=0}^{l={j-1}}{(U^{n{i^{l}}}X_{1}^{n_{1}{i^{l}}}+U^{n{i^{l}}}X_{2}^{n_{2}{i^{l}}}+...+U^{n{i^{l}}}X_{i}^{n_{i}{i^{l}}})})^{-(i-1)}}

= U i ( i − 1 ) i j X k i ( i − 1 ) i j ( ∏ l = 0 l = j − 1 ( U i ( i − 1 ) i l X 1 i ( i − 1 ) i l + . . . + U i ( i − 1 ) i l X i i ( i − 1 ) i l ) ) − ( i − 1 ) {\displaystyle =U^{i(i-1){i^{j}}}X_{k}^{i(i-1){i^{j}}}(\prod _{l=0}^{l={j-1}}{(U^{i(i-1){i^{l}}}X_{1}^{i(i-1){i^{l}}}+...+U^{i(i-1){i^{l}}}X_{i}^{i(i-1){i^{l}}})})^{-(i-1)}}

= U ( i − 1 ) i j + 1 X k ( i − 1 ) i j + 1 ( ∏ l = 0 l = j − 1 ( U ( i − 1 ) i l + 1 x 1 ( i − 1 ) i l + 1 + . . . + U ( i − 1 ) i l + 1 X i ( i − 1 ) i l + 1 ) ) − ( i − 1 ) {\displaystyle =U^{(i-1)i^{j+1}}X_{k}^{(i-1)i^{j+1}}(\prod _{l=0}^{l={j-1}}{(U^{(i-1){i^{l+1}}}x_{1}^{(i-1){i^{l+1}}}+...+U^{(i-1){i^{l+1}}}X_{i}^{(i-1){i^{l+1}}})})^{-(i-1)}}

It is the expression of  x k , j + 1 {\displaystyle x_{k,j+1}} of the exponent i-1.

If we suppose that exist solutions for the exponent i-1,

there will exist solutions for an exposant not greater than i(i-1).

Some times, we must make attention to the initial change of the data. For example, let the following equationd

k U n = X n + Y n {\displaystyle kU^{n}=X^{n}+Y^{n}}

for some k integers like 7, there are solutions, for others like 2, there is no solution. It is too easy toi pose

u = ( k U n ) 2 {\displaystyle u=(kU^{n})^{2}}

x = k U n X n {\displaystyle x=kU^{n}X^{n}}

y = k U n Y n {\displaystyle y=kU^{n}Y^{n}}

z = X n Y n {\displaystyle z=X^{n}Y^{n}}

the lemma 1 is satisfied

u = k U n ( X n + Y n ) = x + y {\displaystyle u=kU^{n}(X^{n}+Y^{n})=x+y}

1 z = k 2 U 2 n k U n X n k U n Y n = u x y = 1 x + 1 y {\displaystyle {\frac {1}{z}}={\frac {k^{2}U^{2n}}{kU^{n}X^{n}kU^{n}Y^{n}}}={\frac {u}{xy}}={\frac {1}{x}}+{\frac {1}{y}}}

The correct solution is to pose

u = U 2 n {\displaystyle u=U^{2n}}

x = U n X n {\displaystyle x=U^{n}X^{n}}

y = U n Y n {\displaystyle y=U^{n}Y^{n}}

z = X n Y n {\displaystyle z=X^{n}Y^{n}}

Like this

u = U 2 n = k U n ( X n + y n ) k 2 = x + y k {\displaystyle u=U^{2n}={\frac {kU^{n}(X^{n}+y^{n})}{k^{2}}}={\frac {x+y}{k}}}

and

1 z = U 2 n U n X n U n Y n = u x y = x + y k x y = 1 k x + 1 k y {\displaystyle {\frac {1}{z}}={\frac {U^{2n}}{U^{n}X^{n}U^{n}Y^{n}}}={\frac {u}{xy}}={\frac {x+y}{kxy}}={\frac {1}{kx}}+{\frac {1}{ky}}}

## Conclusion

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The conclusion is that Ghanouchi's sequences, series and numbers have several applications in all diophantine equations, we saw some of them and there are many others like Pilai equation, Smarandache equation, the Catalan equation, etc...