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

$$\overbrace{(x+y)^4}^A-4\overbrace{xy^2(3x+2y)}^B=1$$
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
$$(x,y)\;=\;(0,\pm 1),\; (\pm1,0),\; (\pm2,\mp3),\; (\pm3,\pm2)\quad\blacksquare $$

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