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.

$a^2+b^2=2c^2\tag{1}$
We can get infinitely many integer solutions by Diophantus's method below.
Substitute  $a=t+p, b=t+q, c=t+r$  to equation $(1)$, then we get 
$$(2p+2q-4r)t+p^2-2r^2+q^2=0$$ 
Let $t = -1/2(p^2-2r^2+q^2)/(p+q-2r)$.
Then we get a parametric solution below.
$a = p^2+(-4r+2q)p+2r^2-q^2$
$b = p^2-2r^2-q^2-2pq+4qr$
$c = p^2+2r^2+q^2-2pr-2qr$
p,q,r are arbitrary.
Thus, we get infinitely many integer solutions.  
Example:  
                     [p,q,r]    [a,b,c]  

                     [3, 2, 1], [7, 1, 5]
                     [4, 3, 1], [17, 7, 13]
                     [5, 2, 1], [23, 7, 17]
                     [5, 4, 1], [31, 17, 25]
                     [6, 3, 1], [41, 1, 29]
                     [6, 5, 1], [49, 31, 41]
                     [7, 2, 1], [47, 23, 37]
                     [7, 6, 1], [71, 49, 61]

