If $x,y$ are Positive integers such that $3x^2+x=4y^2+y$ Prove that $x-y$ is a Perfect Square
My try:
We have $$3x^2+x-(4y^2+y)=0$$ a Quadratic in $x$
So $$x=\frac{-1+\sqrt{1+12(4y^2+y)}}{6}$$
This implies $48y^2+12y+1$ should be a perfect Square say $m^2$
$$48y^2+12y+(1-m^2)=0$$
Whose Discriminant is $144-192(1-m^2)=192m^2-48$ should be a Perfect Square say $p^2$
So
$$192m^2-48=p^2$$
Any way further?
