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.

It is equivalent to solving the circle equation $x^2 + y^2 = 2$ in rational numbers.
Take one point $(1,1)$ on this curve and consider the line with slope $t$ passing through that line: $y = t x - t + 1$, substitute this in $x^2 + (t x - t + 1)^2 - 2 = 0$ and divide out $(x-1)$ to find the second point of intersection between this line and the circle, $(t^2 + 1)x + (-t^2 + 2t + 1) = 0$.
We have parametrized all nontrivial solutions $$x = a/c = \frac{-t^2 + 2t + 1}{t^2 + 1}$$
For example $t = 22/7$ gives us $a = 127$, $c = 533$ and $127^2 + 743^2 = 2\cdot 533^2$.
