Diophantine Equations : Solving $a^2+ b^2=2c^2$

Assume you have the Pythagorean relation $u^2 + v^2 = c^2$
Then $$ \begin{align} (u^2 + v^2) + (u^2 + v^2) & = 2c^2\\ (u^2 + v^2 + 2uv) + (u^2 + v^2 - 2uv) & = 2c^2\\ (u + v)^2 + (u - v)^2 & = 2c^2\\ \end{align} $$ Thus if $a = u + v$ and $b = |u - v|$
$a^2 + b^2 = 2c^2$

Update

Given any triple of integers $a, b, c : a^2 + b^2 = 2c^2$
$a^2 + b^2 \equiv 0 \mod 2$
Which implies $a + b \equiv 0 \mod 2$,
So $a \equiv b \mod 2$ and
$a - b \equiv 0 \mod 2$

WLOG, assume $a \ge b$
Both $a + b$ and $a - b$ are even.
Let $2u = a + b$ and $2v = a - b$
$4u^2 + 4v^2 = (a + b)^2 + (a - b)^2 = 2(a^2 + b^2) = 4c^2$
$u^2 + v^2 = c^2$
i.e., $u, v, c$ is a Pythagorean triple.

So every triple of integers $a, b, c : a^2 + b^2 = 2c^2$ corresponds to a Pythagorean triple.

Hint: In $\mathbb Z[i]$, let $x:=a+bi$, $y:=c$, then $x\bar x=(1+i)(1-i)y^2$. $\mathbb Z[i]$ being UFD implies that $1+i\mid x$ or $1-i\mid x$. If $1+i\mid x$, write $z:=\frac x{1+i}=\frac {x(1-i)}2=\frac{a+b}2+\frac{b-a}2i$. Then we have $(\frac{a+b}2)^2+(\frac{b-a}2)^2=z\bar z=y^2=c^2$.

Here is a geometric way to look at the posted answers: Integer solutions to $a^2+b^2=c^2$ correspond to ${\mathbb Q}$-points $(a/c,b/c)$ on the circle $S$ defined by $x^2+y^2=1$. Integer solutions to $a^2+b^2=2c^2$ correspond to ${\mathbb Q}$-points $(a/c,b/c)$ on the circle $T$ defined by $x^2+y^2=2$. Let $r:=\sqrt{2}$. There are various correspondences between the circles $S$ and $T$, the most obvious being $(x,y)\mapsto(rx,ry):S\to T$. The problem is that the coefficients being used in this map are not all elements of ${\mathbb Q}$, so this map does not send ${\mathbb Q}$-points to ${\mathbb Q}$-points. The observation being made in several of the comments above is that, if we precompose this map with the rotation $(x,y)\mapsto((x-y)/r,(x+y)/r):S\to S$, then the resulting map $(x,y)\mapsto(x+y,x-y):S\to T$ $is$ defined over ${\mathbb Q}$ ($i.e.$, has coefficients in ${\mathbb Q}$), and so it will carry ${\mathbb Q}$-points to ${\mathbb Q}$-points. This then gives a correspondence between integer solutions to $a^2+b^2=c^2$, aka Pythagorean triples, and integer solutions to $a^2+b^2=2c^2$. This brings up the question of when two ${\mathbb Q}$-conic sections are ${\mathbb Q}$-isomorphic, something that may well be known, but not by me.