A Square Expression

[sowmitra]

Find all ordered pairs $(a,b)$ of positive co-prime integers such that,
$\displaystyle a^2-6ab+b^2$ is a perfect square.

[sourav das]

W.L.O.G. let $a>b$. Let $a^2-6ab+b^2=2(a-b)^2-(a+b)^2=c^2$. Now, substitute $a-b=p$...(i), $a+b=p+2gy$....(ii), $c=p-2gx$...(iii) with $x,y$ co-prime. Then, you will find out that $a-b=p=g(x-y)+ \frac{2gxy}{x-y}$. Using (i),(ii) we will get $a=gx+\frac{2gxy}{x-y}$ and $b=gy$ . But since $(xy,x-y)=1$ as $(x,y)=1$; so, $2g=(x-y)k$ for some $k\in \mathbb N$. So we will get,
$a=\frac{(x-y)xk}{2}+xyk$ and $b=\frac{(x-y)yk}{2}$ But since $(a,b)=1$, so $k\in \{1,2\}$
Case(i): $k=2$ then $a=x(x+y)$ and $b=y(x-y)$ also note that $(x,y)=1$ and only one of $x,y$ must be even in this case.
Case(ii): $k=1$ then $a=\frac{x(x+y)}{2}$ and $b=\frac{y(x-y)}{2}$. Here $(x,y)=1$ and both of them must be odd.

Check:Just note that $x^2(x+y)^2-6xy(x^2-y^2)+y^2(x-y)^2=(x^2-2xy-y^2)^2$

[Tahmid Hasan]

Now, substitute $a-b=p$...(i), $a+b=p+2gy$....(ii), $c=p-2gx$...(iii) with $x,y$ co-prime.

How did you make these substitutions? Some insights would really help.

[sowmitra]

Tahmid-এর সাথে একমত। ভাইয়া, Substitution-গুলো একটু Explain করলে ভালো হত...

[sourav das]

I just wanted to solve viewtopic.php?f=27&t=2568&p=12849#p12849
and wanted to represent the smallest number( $c$ ) with something else. First $p+2y$ and then $p+2gy$ with co-prime condition.