2020 Mathematics Subject Classification: Primary: 11A07 [MSN][ZBL]
Let $p$ be a prime number greater than 3. The numerator of the fraction $$ \frac{1}{1} + \frac{1}{2} + \cdots + \frac{1}{p-1} $$ is divisible by $p^2$.
An equivalent form of the theorem is that if $x^*$ denotes the solution to the equation $x x^* \equiv 1 \pmod {p^2}$ then $$ 1^* + 2^* + \cdots + (p-1)^* \equiv 0 \pmod{p^2} \ . $$