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.
 A: 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...
A: Noting that
$a^2+b^2=2c^2 \Rightarrow a $ and $b$ are of same parity. Hence there exists natural numbers $u$ and $v$ such that
$$a+b=2u \textrm{ and } a-b=2v$$
Then $$a=u+v \textrm{ and }b=u-v$$
$$
\begin{gathered}
(u+v)^2+(u-v)^2=2 c^2 \Rightarrow 
u^2+v^2=c^2
\end{gathered}
$$
There exists Pythagorean triple such that
$$
\begin{aligned}
& \left\{\begin{array}{l}
u=2k m n \\
v=k(m^2-n^2) \\
c=k(m^2+n^2)
\end{array}\right. \Rightarrow  \left\{\begin{array}{l}
a=k(2 m n+m^2-n^2 )\\
b=k(2 m n-m^2+n^2) \\
c=k(m^2+n^2)
\end{array}\right. ,\\
&
\end{aligned}
$$
where $k,m,n \in N$.
A: Here is a proof on why natural solutions exist for $a, b, c$, as a matter of fact, infinite solutions.
$$a^2 + b^2 \equiv 0 \mod 2 \to (a + b)^2 - 2ab \equiv 0 \mod 2  \\ \quad \\ \implies a + b \equiv 0 \mod 2 $$
More importantly, this suggests $$a \equiv b \mod 2 \implies \text{ both $a - b$ and $a + b$ must be even}$$
Then there are such integers $x, y$ which satisfy $a - b = 2x$ and $a + b = 2y$.
Hence,
$$4x^2 + 4y^2 = (a + b)^2 + (a-b)^2 = 2(2c^2) \\ \quad \\ \implies x^2 + y^2 = c^2 * \qquad \square$$

For better understanding, $a, b$ have to be both even or odd, hence have the same remainder when divided by $2$. Furthermore, $*$ suggests there are infinite solutions to $a^2  + b^2 = c^2$ as there are infinite integer pythagorean triples.
