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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?

Michael Rozenberg
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Anonymous Pi
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2 Answers2

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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 $ab=mn$.

Michael Rozenberg
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    Followed maybe by a substitution? I feel like this is a very vague lead. – Anonymous Pi Sep 16 '17 at 02:07
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    @ Anonymous P Now we can assume that $(a,b)=1$ and there are $m$ and $n$ with $(m,n)=1$ and different parity for which $ab=mn$ and $a^2-2ab-b^2=m^2-n^2$. – Michael Rozenberg Sep 16 '17 at 02:14
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    I suspect that what Michael has in mind is that in order for $P(a,b)$ to be a square value, the rewrite above puts it into the much smaller category of a Pythagorean triple, as also the sum of two square values. – hardmath Sep 16 '17 at 02:37
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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 $b=0$.

Allan MacLeod
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