## Task A-4.7.

Determine all pairs of natural numbers $(x, y)$ for which $x^{2}-y!=2016$.
## First solution.

The number 2016 is divisible by 32 and is not divisible by 64.

Notice that it cannot be the case that $y=1$. Therefore, it must be that $y \geqslant 2$, so $y$ is even. Hence, $x$ must be an even number.

Since $x^{2}$ and 2016 are both divisible by 4, $y$ must also be divisible by 4, so $y \geqslant 4$. 1 point

Now we can conclude that $y$ is divisible by 8, so $x^{2}$ must be divisible by 8. This means that $x$ must be divisible by 4, and thus $x^{2}$ is divisible by 16. Therefore, $y$ is divisible by 16, so $y \geqslant 6$.

If $y \geqslant 8$, then $y$ is divisible by 128, and particularly by 32, so $x^{2}$ must be divisible by 32.

By checking, we see that $2016+6$ is not a perfect square, but $2016+7!=7056=84^{2}$, so the only solution is $(x, y)=(84,7)$.