TITLE: Diophantine Equations : Solving $a^2+ b^2=2c^2$
QUESTION [16 upvotes]: 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 ...

REPLY [2 votes]: 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.