How many pairs are there?

We may tackle the problem by setting up the following system of equations: $x^2 + 3y = (x+t)^2 \\ y^2 + 3x = (y+c)^2$ For some positive integers $t,c$ . The good thing about this setup is that we can actually bound the values at least one of these variables. Suppose that both $t,c > 1$ , then $t,c \geq 2$ and thus $x^2 +3y \geq (x+2)^2 \implies 3y \geq 4x + 4 \\ y^2 + 3x \geq (y+2)^2 \implies 3x \geq 4y + 4$ Adding these two inequalities we obtain $3(x+y) \geq 4(x+y) + 8$ , a clear contradiction. Without loss of generality let's now assume that $t=1$ . We may do this because it is also impossible to have $t < 1$ . This immediately tells us $x^2 + 3y = (x+1)^2 \implies 3y = 2x + 1$ . We could now just plug this into the various expressions we have, but I found it useful to clean this up a little bit. The given equation implies that $2x + 1 \equiv 0 \mod 3 \implies x \equiv 1 \mod 3 \implies x = 3k + 1$ for some non-negative integer $k$ . This also tells us $y = 2k + 1$ .

Because we have the identity $x^2 + 3y = (x+1)^2$ , we already know this expression is a perfect square. We can now just focus on figuring out when is $y^2 + 3x$ a perfect square. Substituting for $y,x$ the expressions with $k$ we find that $4k^2 + 13k + 4$ must be a perfect square. Notice that $4k^2 + 13k + 4 = (2k + 3)^2 + k - 5$ . If $k > 5$ , then we would have $(2k + 3)^2 + k - 5 \geq (2k + 4)^2 \implies k-5 \geq 4k + 7 \implies 0 \geq 3k + 12$ , a contradiction. Thus, we just have to check the values $0 \leq k \leq 5$ . Doing this, we find that the polynomial is only a perfect square when $k=0,5$ . These give the pairs $(1,1), (11,16), (16,11)$ .