<s>[INST] Diophantine Equations : Solving $a^2+ b^2=2c^2$ I was working through some number theory problems , when I came across the following question :

Find all solutions of $a^2+b^2=2c^2$

My Solution (Partial) :

  
*
  
*We can rewrite the above equation as : $c^2 = (a^2 + b^2)/2 $
  
*So $\Rightarrow$ $a^2 , c^2 , b^2$ are in an Arithmetic Progression $\Rightarrow$ there exist infinite solutions
  
*WLOG , let $a^2$ be $t$ , $b^2$ be $s$ and $c^2$ be $m$
  
*The equation can be re-written as $ t + s = 2m$
  
*Now , what I was thinking was that should I solve this Diophantine equation for $t$ & $s$ in terms of $m$ ; substitute their values back into the equation and find out $m$ ; and then finally put this value of $m$ back into the values of $t$ and $s$ to derive a general solution
  

Can someone help me out ? Maybe a hint ...
 [/INST] 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.
</s>