<s>[INST] Does $a^2 + b^2 = 2 c^2$ have any integer solution? 
Does the equation $a^2 + b^2 = 2 c^2$ have any integer solution with $|a| \neq |b|$?

I think not, because of these equations for pythagorean triplets:
$$\left(\frac{a}{c}\right)^2+\left(\frac{b}{c}\right)^2=1$$
$$x=\frac{1-t^2}{1+t^2}, y=\frac{2t}{1+t^2}$$
I think I would need to multiply $x$ and $y$ by $\sqrt{2}$ and they would never be rational numbers.
 [/INST] Suppose $(x,y,z)$ is any Pythagorean triple. Then:
$$(x-y)^2+(x+y)^2=2x^2+2y^2=2z^2$$
</s>