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 $b = |u - v|$
$a^2 + b^2 = 2c^2$

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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 $(\frac{a+b}2)^2+(\frac{b-a}2)^2=z\bar z=y^2=c^2$.