Mathematical Reflections 2014, Issue 1 - Problem S290

Problem:
Prove that there is no integer $n$ for which $$\dfrac{1}{2^2}+\dfrac{1}{3^2}+\ldots+\dfrac{1}{n^2}=\left(\dfrac{4}{5}\right)^2.$$

Proposed by Ivan Borsenco.

Solution:
Let $p$ be the greatest prime number such that $p \leq n < 2p$. Then the given equality can be written as $$5^2k=(4\cdot n!)^2,$$ where $k=\displaystyle \sum_{i=1}^n \dfrac{(n!)^2}{i^2}$. Observe that $k \equiv \dfrac{(n!)^2}{p^2} \not \equiv 0 \pmod{p}$. Since $p|5^2k$ and $p$ does not divide $k$, it follows that $p|5^2$, i.e. $p=5$. So, $n \in \{5,6,7,8,9\}$. An easy check shows that none of these values satisfies the equality.