Question

Find the general values of $x$ and $y$ which make the following expression a perfect square:
$x^2-3xy+3y^2$.

Answer 1

$x^2-3xy+3y^2=z^2$ (say)

$x^2-z^2=3y(x-y)$

Put $m(x+z)=3xy$ and

$n(x-z)=m(x-y)$

then by cross-multiplication;-

$\frac{x}{3n^2-m^2}=\frac{y}{2mn-m^2}=\frac{z}{m^2-3mn+3n^2}$

Hence, $x=3n^2-m^2,y=2mn-m^2$ and the hypotenuse is $m^2-3mn+3n^2$.