Does $a^2 + b^2 = 2 c^2$ have any integer solution?

There are infinitely many solutions with $a=1$. We want to solve $b^2-2c^2=-1$. Armed with the solution $b=c=1$ and the fact that $3^2-2\cdot 2^2=1$ we can use the Brahmagupta-Fibonacci identity to find more. Given a solution $(i,j)$ the next one is $(3i+4j,2i+3j)$, leading to the sequence $(7,5), (41,29),(239,169),(1393,985)\ldots$ You can negate either or both members of these pairs because $b$ and $c$ are squared.

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 $127^2 + 743^2 = 2\cdot 533^2$.

Suppose $(x,y,z)$ is any Pythagorean triple. Then: $$(x-y)^2+(x+y)^2=2x^2+2y^2=2z^2$$