Solving $x^3+y^3=x^2y^2+1$ in non-negative integers

Question

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!

Answer 1

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.)

Answer 2

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.