So, I randomly thought about this simple problem, and very sadly couldn't make much progress in answering this.
Does there exist three integers $a,b,c\neq0$ where $a^2+b^2$, $a^2+c^2$, and $b^2+c^2$ are all perfect squares?
Here's what I've worked out:
Let $a^2+b^2=C^2$,
$a^2+c^2=B^2$,
$b^2+c^2=A^2$.
- For the smallest $a,b,c$ triplet that exists, exactly one of them must be odd.
- It wouldn't be the smallest triplet if they're all even, and...
- ...there doesn't exist a Pythagorean triple $x^2+y^2=z^2$ where both $x$ and $y$ are odd (so $a,b,c$ can't have more than one odd number).
- I've also found that $2(a^2+b^2+c^2)=A^2+B^2+C^2$, but this makes sense even when one of $a,b,c$ is odd.
I've even tried brute-forcing this with some code, but to no avail. So, I'm guessing there's no such $a,b,c$, but I haven't figured out how to prove that
So, it does seem that answering this would imply whether $a,b,c$ exists:
Can $b^2+c^2$ be a perfect square if $a^2+b^2$ and $a^2+c^2$ are?
...which I haven't figured out either.
Maybe some smart people have already thought about this many years ago – if so, I'd be happy to see their work! (Though, I've searched this up and found nothing.)
Also, feel free to add/fix the tags (I only added two here). Thanks in advance!
Such a system of Diophantine equations arises from so-called Euler bricks, which are well known. The smallest solution is given by \begin{align*} 44^2+ 117^2 & = 125^2,\\ 44^2+ 240^2 & = 244^2,\\ 117^2+240^2 & = 267^2. \end{align*}