All natural number solutions for the equation $a^2+b^2=2c^2$ $a$, $b$ and $c$ of all Pythagorean triplets can be written in the form
$$
\begin{split}
a &= 2mn\\
b &= m^2-n^2 \\
c &= m^2+n^2
\end{split}
$$
where $m$ and $n$ are natural numbers. For any natural number $m$ and $n$, this set of equations will give a Pythagorean triplet. And all Pythagorean triplets satisfy this set of equations.
Can $a$, $b$ and $c$ of all triplets satisfying the equation $$a^2+b^2=2c^2$$ where $a$, $b$ and $c$ are natural numbers, be written as a set of equations as for the Pythagorean triplets?
So, I need a set of equations that generates triplets that satisfy the equation $a^2+b^2=2c^2$ for any natural numbers I plug into the set of equations. Also, every natural number triplets satisfying the equation $a^2+b^2=2c^2$ must satisfy the set of equations.
I tried to derive the set of equations myself, no attempts have been successful yet.
I would like to have the proof of the set of equations, (otherwise I won't know if every triple will satisfy the set of equations)
Any comments that helps to give an insight into solving the problem are really appreciated.

Hint: For $a^2+b^2=2c^2$, observe that $a, b$ have the same parity. Therefore there exist integers $u, v$ such that $a = u+v$ and $b = u-v$. Expand...
