I got this question from RMO-2014
Find all $n\in\mathbb N$ such that $n^2-440$ is a perfect square.
My attempt:
Let $n^2-440=(n-k)^2$
Then $k^2-2nk+440=0$
This gives $k=\frac{2n\pm2\sqrt{n^2-440}}{2}$, which gives $n=k$, which gives $n\not\in\mathbb N$ or $k=k$, which is useless.
$21$ is an answer, by trial and error, but I can't go any further.
Please help.
Let $n^2-440=m^2$.
Then $(n+m)(n-m)=440$
$440=2^3\cdot5\cdot11$.
Now, $440$ can be represented as
$\begin{align}1&\times440\\2&\times220\\4&\times110\\5&\times88\\8&\times55\\10&\times44\\11&\times40\\20&\times22 \end{align}$
So we have if $n+m=a,n-m=b$, we have $n=\frac{a+b}2$. This is integral only when, in this case, $a,b$ are both even.
So we get $n=21,27,57,111$.