On the square values of a certain polynomial

Question

Consider the following polynomial on two variables: $$P(a,b)=a^4-4a^3b+6a^2b^2+4ab^3+b^4.$$ Do positive integers $x$ and $y$ exist such that $P(x,y)$ is a perfect square?

I'm aware that this may be actually a very hard problem in disguise, since I'm trying to prove something that I'm not aware that has been proven yet by using this as a lemma, and since this is number theory after all. However, the striking resemblance of this polynomial to $(a\pm b)^4$ and the fact that its homogeneous make me believe there may be a not-so-hard way to prove/disprove this conjecture. I've tried to limit the expression between two consecutive squares, I've tried module and I even tried to gather some evidence: No solutions with $$10000>a>b$$ exist. Is there something I'm missing? Can more heavy machinery be used?

Answer 1

Try to use $$a^4-4a^3b+6a^2b^2+4ab^3+b^4=(a^2-2ab-b^2)^2+4a^2b^2.$$ Now, we can assume $(a,b)=1$ and there are $m$ and $n$ with different parity such that $(m,n)=1$

for which $a^2-2ab-b^2=m^2-n^2$ and $ab=mn$.

Answer 2

The answer is NO - the only integer solutions involve $a=0$ or $b=0$.

The quartic is birationally equivalent to the elliptic curve \begin{equation} v^2=u^3+3u^2+u \end{equation} with \begin{equation} \frac{a}{b}=\frac{u+v}{u+1} \end{equation}

Using Pari-GP's inbuilt elliptic curve functions and Denis Simon's ellrank software, it is straightforward to show that the only rational points on the curve are $(0,0)$ and $(-1, \pm 1)$. These only lead to $a=0$ or $b=0$.