A Take on Pythagorean-Triples

[sowmitra]

Find all ordered-pairs $(a,b,c)$ of co-prime positive integers, such that,
\[\displaystyle a^2+b^2=2c^2\]

[sourav das]

$2c^2-a^2=b^2$ (assuming $a>b$) and now substitute $a=x+y$ and $c=x-y$ and our problem transform into,
viewtopic.php?f=27&p=12845&sid=f2621629 ... 17d#p12845

[Tahmid Hasan]

I've assumed $(a,b,c)$ are pairwisely co-prime.
I solved the problem with Sourav's substitution
$a^2+b^2=2c^2$.....$(1)$
WLOG $a>b$, then $a>c>b$.
Let $b=c-2gx,a=c+2gy;(x,y)=1,g \in \mathbb{N}$.
Then plugging the values in $(1)$ implies $g=\frac {c(x-y)}{x^2+y^2}$.
Hence $b=c. \frac {y^2+2xy-x^2}{x^2+y^2},a=c. \frac {x^2+2xy-y^2}{x^2+y^2}$.
Since $(x,y)=1$ we get $(x^2+y^2,y^2+2xy-x^2)=1;(x^2+y^2,x^2+2xy-y^2)=1$ which implies $x^2+y^2 | c$.
But since $(a,b)=1$, we conclude $c=x^2+y^2$.
So $a=x^2+2xy-y^2,b=y^2+2xy-x^2$.
Identity $(x^2+2xy-y^2)^2+(y^2+2xy-x^2)^2=2(x^2+y^2)^2$