More fun in 2016, Part 4

Let $\begin{cases}n=k^2\\n+2016=s^2\\n-2016=t^2\end{cases}$ . so we find that $s^2-k^2=2016\Longrightarrow (s-k)(s+k)=2016$ $k^2-t^2=2016\Longrightarrow (k-t)(k+t)=2016$ We see that $s - k$ and $s + k$ are both factors of 2016. Similarly for $k-t$ and $k+t.$ This gives us a simultaneous system of equations for which we're looking for integer solutions. If we let the solutions to $a^2-b^2=2016$ be $(a_1,b_1),(a_2,b_2)...,$ then we must have $s$ as an $a_i$ and $t$ as a $b_i$ . However, $k$ must also be both an $a_i$ and a $b_i$ . We find all pairs in a decreasing order and stop once we find the potential $k$ . We check and see that $k=65$ satisfies the equation and is indeed the biggest possible $k$ . The answer is hence $n=k^2=4225$

This is an example of the Congruum Problem discussed and solved by Fibonacci in his book "Liber Quadratorum" of 1225.

What follows is merely an outline of a solution.

The problem is closely related to Pythagorean triples. Indeed, if we have a triple $a^2=b^2+c^2$ with $bc=1008$ , then $a^2\pm 2bc=a^2 \pm 2016=(b\pm c)^2$ , so that $n=a^2$ solves our problem.

To find a triple $a^2=b^2+c^2$ with $bc=1008$ , we use Euclid's formula and write $b=2pq$ and $c=p^2-q^2$ . From the condition $bc=1008$ we quickly find that $p=8,q=1, b=16, c=63$ and $n=a^2=b^2+c^2=\boxed{4225}$

We can also figure out that the last digit of the number of which n is square must be 5. Now its easy to check with calculator in hand that 65 is the solution