Solving $x^3+y^3=x^2y^2+1$ in non-negative integers I wanted to solve $x^3+y^3=x^2y^2+1$ in non-negative integers.
First I set $a=x+y$ and $b=xy$ to get $b^2+3ab+1=a^3$. View as a quadratic in $b$, the discriminant = $4a^3+9a^2-4$, which needs to be a perfect square.
Secondly, rearranging the quadratic in $b$ we get $4a^3+9a^2-4=(2b+3a)^2$.
So the discriminant is always a perfect square. Therefore we have (quadratic formula):
$b=\frac{-3a\pm (2b+3b)}{2}$ so $b\in \{b,-\frac{3a}{2}\}$.
Since we want $a,b\ge 0$, the only possibility is $a=b=0$ to give $x=y=0$.
This is the unique solution.
Note: I wasn't sure if it works, I never tried this way before.
Thanks!

I think the following fact has been violated:-
In the quadratic equation, f(X) = 0, its discriminant ($\triangle$) should be free from X.
It is alright to have $\triangle = 4a^3 + 9a^2 – 4$, but rewriting it as $(2b + 3a)^2$ and use it to solve f(b) = 0 is not allowable.
