Proofathon Squares!

We claim that there is no such triple. We're going to prove this by contradiction.

Let,

$a^2+7b^2+5=x^2$ $b^2+7c^2+5=y^2$ $c^2+7a^2+5=z^2$ where $x, y, z \in \mathbb{Z}$ .

For some reason, it seems like adding them up is a good idea, doesn't it?

Do that to get $8(a^2+b^2+c^2)+15=x^2+y^2+z^2$ .

Notice that the left hand side is congruent to $-1$ modulo $8$ .

But since perfect squares can only be congruent to $0, 1$ and $4$ modulo $8$ , the right hand side can not be congruent to $-1$ modulo $8$ .

[This is just case-checking. The possible residues the right hand side can give modulo $8$ are $0, 1, \pm 2, \pm 3, 4$ . $-1$ isn't one of them].

That means there can not exist a triplet of integers $(a, b, c)$ that satisfies our conditions and our answer is $\boxed{0}$ .