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While working on some research, I needed to find the integer solutions to the polynomial $x^4+4x^3y-6x^2y^2-4xy^3+y^4=1$. I found $(0,1)$, $(0,-1)$, $(1,0)$, $(-1,0)$, $(3,2)$, $(-3,-2)$, $(2,-3)$, $(-2,3)$. These seem to be all, but can I prove this? I factored the equation several ways. I have: $(x+y)^4-4xy^2(3x+2y)=1$, $(x-y)^4-4x^2y(3y-2x)=1$, and $x^4+2xy(2x+y)(x-2y)+y^4=1$. Any suggestions? Even if I could find some bounds for possible solutions, that would be great.

user539887
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vrt
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    You can use MathJax in the title, too. – user539887 May 01 '18 at 22:14
  • Take the form $(x+y)^4-4xy^2(3x+2y)=1$, suppose $x+y=T$ you get a Pell like equation as $(T^2)^2 -4x(3x+2y) y^2=1$, compare it with pell equation $x^2- D y^2=1$ which can have infinitely many distinct integer solutions. $D=4x(3x+2y)$ can have particular values. – sirous May 02 '18 at 01:35
  • For $|x|,|y|\le 10^4$, no further solutions exist. – Peter May 02 '18 at 09:13
  • According to the thue-command in PARI/GP, the list is complete. – Peter May 02 '18 at 09:16
  • Isn't there a limit to the number of solutions to a Pell-like equation if x is a perfect square? That is, isn't there a finite number of solutions to $x^4-dy^2=1$? – vrt May 03 '18 at 20:33
  • @vrt The main problem with the Pell-approach is that $D$ depends on $x$ and $y$, which is not the case in a usual Pell-equation. So I don't think such an approach leads to anywhere. – Peter May 06 '18 at 15:58
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    @Peter I realised the same thing. But you've been helpful, as have the others who posted. Thank you! – vrt May 06 '18 at 19:07

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