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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 ...

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    $\begingroup$ There are infinitely many solutions. See here (end of the 9th page). $\endgroup$ – user26486 Apr 25 '15 at 7:01
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    $\begingroup$ If the coefficient can be represented as a sum of squares, the formula there. artofproblemsolving.com/community/… $\endgroup$ – individ Apr 25 '15 at 7:18
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    $\begingroup$ I think question was good, so +1 :) $\endgroup$ – Sawarnik Apr 26 '15 at 18:15
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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.

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    $\begingroup$ For the sake of completeness, I point out this answer only shows that you can construct solutions out of pythagorean triples, and doesn't show that all solutions come from pythagorean triples. $\endgroup$ – Hurkyl May 29 '15 at 16:00
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    $\begingroup$ @Hurkyl: Fair comment. The OP said they wanted a hint, so I didn't post a complete answer originally. I guess there's no harm in adding a little more detail now, but I won't bother showing how to generate all Pythagorean triples, since that's already well-documented. $\endgroup$ – PM 2Ring May 30 '15 at 7:20
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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 $(\frac{a+b}2)^2+(\frac{b-a}2)^2=z\bar z=y^2=c^2$.

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