2016 and factorial

We consider two cases.

Case 1 : $m \leq 7$ .

This gives us $8$ cases to check and shows the only solution to be $7!+2016=7056=84^2$ .

Case 2 : $m \geq 8$ .

$m \geq 8 \implies 64 \vert m! \implies m!=64 k \quad k \in \mathbb{N} \quad \quad \color{#3D99F6}{\text{As } (2 \times 4 \times 8) \vert 8!}$

$m!+2016=64k+2016=16(4k+126)=\left (4^2 \right) \left (4k+126 \right) \implies 4k+126=n^2 \quad n \in \mathbb{N}$

We know $n^2 \equiv 0 \text{ or } 1 \pmod{4}$ but $4k+126 \equiv 2 \pmod{4}$ which is a contradiction so there are no solutions $m \geq 8$ .

This means the only solution is $m=7$ giving the sum of the perfect squares as $\color{#20A900}{\boxed{\boxed{7056}}}$

Suppose that $m!+2016=n^2$ with $n$ is a positive integer.

If $m\le 7$ it is easy to check that $m=7$ is the unique solution. In that case, we have: $7!+2016=7056=84^2$

If $m\ge8$ , we have $m!+2016\equiv 32\ (\bmod{64})$ , cannot be a perfect square.