Perfect Square!!

Let $\begin{aligned} N^2 & =\binom{n}{3}\binom{n}{4}\binom{n}{5}\binom{n} {6} \\& =\left( \dfrac{\,(n!)^4}{2^{11}\cdot 3^5 \cdot 5^2 \,(n-3)!(n-4)!(n-5)!(n-6)!}\right)^{-2} \\ & =\dfrac{\,(n!)^2}{2^5\cdot 3^2 \cdot 5}\left( \dfrac{1}{6 \,(n-3)((n-4)!)^2(n-5)\,( (n-6)!)^2}\right)^{-2} \\ & =\dfrac{\,(n!)^2}{480 \,(n-3)\,(n-4)!\,((n-6)!)}\left( \dfrac{1}{6 \,(n-3)(n-5) }\right)^{-2} \end{aligned}$ Notice that for $n\geq 6$ , $N$ is perfect square number if $6(n-3)\,(n-5)$ is a perfect square number or $6(n-3)(n-5)$ must be in the form of $36k^2$ ie $6(n-5)(n-3) =36k^2\implies (n-3)(n-5) = 6k^2 \ k\geq 1$ Expanding the Left hand side we have $n^2-8n+15 -6k^2=0\implies n= 4 \pm \sqrt {6k^2+1}$

Giving the possible values of $n=9,53$ for certain values of $k= 2,20$