<s>[INST] 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!
 [/INST] Introducing the variables $a$ and $b$ leads to the condition
$$4a^3+9a^2-4=(2b+3a)^2\ ,$$
which implies that
$$4a^3+9a^2-4$$
has to be a perfect square. A quick computer search ($|a|\leq 10^6$) produced  the solutions
$$a=-2,\quad-1,\quad1,\quad 2,\quad 25\ .$$
This should be some material to work with.
(The logic in the second half of your argument is circular.)
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