Find this diophantine equation intgers $x^4+4x^3y-6x^2y^2-4xy^3+y^4=1$

Question

Find the solutions of the diophantine equation $$x^4+4x^3y-6x^2y^2-4xy^3+y^4=1$$

I have found $(x,y)=(\pm 1,0),(0,\pm 1)$: how to find all solutions?

Thanks

Answer 1

$$\overbrace{(x+y)^4}^A-4\overbrace{xy^2(3x+2y)}^B=1$$ As $x,y$ are integers, a possible solution is $A=1, B=0$ i.e. $$A=1\Rightarrow x+y=\pm 1\quad\cdots (1)\\ B=0\Rightarrow xy^2=0 \Rightarrow x=0 \text{ or } y=0\quad\cdots (2a)\\ \text{or}\\ 3x+2y=0\Rightarrow 3x=-2y\quad\cdots (2b)$$ From $(1),(2a)$, $$(x,y)=(0,\pm1), (\pm1,0)$$ From $(1),(2b)$, $$(x,y)=(2,-3), (-2,3)..$$ Also, another solution is $$(x,y)=(\pm3,\pm2)$$ Hence, integer solutions are $$(x,y)\;=\;(0,\pm 1),\; (\pm1,0),\; (\pm2,\mp3),\; (\pm3,\pm2)\quad\blacksquare $$

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NB: $$\begin{align} 4xy^2(3x+2y)&=(x+y)^4-1\\ &=(x+y-1)(x+y+1)((x+y)^2+1)\\ &=(2n)(2n+2)(4n^2+4n+2)&&(x+y=2n+1)\\ xy^2(3x+2y) &=2n(n+1)(2n^2+2n+1)\\ &=2n(n+1)(2n(n+1)+1)\\ &=u(u+1)&&(u=2n(n+1)) \end{align}$$