Mathematical Reflections 2017, Issue 1 - Problem S400

Problem:
Find all $n$ for which $(n-4)!+\dfrac{1}{36n}(n+3)!$ is a perfect square.

Proposed by Titu Andreescu, University of Texas at Dallas, USA

Solution:
Clearly, $n \geq 4$. We have $$\begin{array}{lll} (n-4)!+\dfrac{1}{36n}(n+3)!&=&(n-4)!\left(1+\dfrac{1}{36n}(n-3)(n-2)(n-1)n(n+1)(n+2)(n+3)\right)\\&=&(n-4)!\left(1+\dfrac{1}{36}(n^2-9)(n^2-4)(n^2-1)\right)\\&=&(n-4)!\left(\dfrac{n^6-14n^4+49n^2}{36}\right)\\&=&(n-4)!\left(\dfrac{n(n^2-7)}{6}\right)^2. \end{array}$$
Observe that $n(n^2-7)=n^3-7n \equiv n^3-n \equiv 0 \pmod{6}$, so $\dfrac{n(n^2-7)}{6}$ is an integer. It follows that $(n-4)!\left(\dfrac{n(n^2-7)}{6}\right)^2$ is a perfect square if and only if $(n-4)!$ is a perfect square. If $n=4$ or $n=5$, then $(n-4)!=1$, which is a perfect square. Let $n>5$ and let $p$ be the greatest prime that divides $(n-4)!$. By Bertrand's Postulate, there exists a prime $q$ such that $p<q<2p$. If $2p \leq n-4$, then $q<n-4$, which gives $q \ | \ (n-4)!$, contradiction. So, $n-4<2p$, which means that $p \ | \ (n-4)!$ and $p^2 \nmid (n-4)!$. So, $(n-4)!$ is not a perfect square if $n>5$. Therefore, $n \in \{4,5\}$.