Find $a$ and $b$ such that $a^2-4b$ and $b^2-4a$ are perfect squares

Question

Under what conditions for $a$ and $b$ are $(1)$ and $(2)$ integers? $$\sqrt{a^2-4b}\tag{1}$$ $$\sqrt{b^2-4a}\tag{2}$$ Apologies, but my only attempt is to show that $a^2-4b$ and $b^2-4a$ are perfect squares, but even on that I have no idea how to approach.

Answer 1

We claim the only solutions are $(4,4)$, $(5,6)$, and $(6,5)$.

Without loss of generality, let $a\leq b$. Then either $b\leq 4$ or

$$(b-3)^2 <b^2-4b \leq b^2-4a \leq b^2.$$

(we have that $(b-3)^2<b^2-4b$ if $b>9/2$.) The case where $b\leq 4$ can be hand-bashed out to reveal only the solution $\boxed{(4,4)}$. Otherwise, as $b^2-4a$ is a square between $(b-3)^2$ and $b^2$, it is either $(b-2)^2$ or $(b-1)^2$. However, we have

$$b^2-4a=(b-1)^2=b^2-2b+1 \implies 4a=2b-1,$$

which is a contradiction since $2b-1$ is odd and thus not a multiple of $4$. So, $b^2-4a=(b-2)^2$, which gives $a=b-1$. Then, we have that

$$a^2-4b=(b-1)^2-4b=b^2-6b+1$$

is a square. However, if $b\geq 15/2$, we have that

$$(b-4)^2=b^2-8b+16 < b^2-6b+1 < b^2-6b+9 = (b-3)^2,$$

a contradiction. Thus, $b\leq 15/2$, reducing the problem to a finite case check revealing only $\boxed{(5,6)}$, and correspondingly $\boxed{(6,5)}$.