<s>[INST] Finding solutions of a 2-variable biquadratic equation Find all integral solutions of $$y^2+y=x^4+x^3+x^2+x$$ Factoring both sides we get $$y(y+1)=(x^2+x)(x^2+1)$$ Let $Y=y+1/2$ and $X=x^2+\frac {x+1}2$. Therefore $$(Y+1/2)(Y-1/2)=(X+\frac{x-1}2)(X-\frac {x-1}2)\\=>Y^2-1/4=X^2-\frac {x^2-2x+1}4\\=>4Y^2-1=4X^2-(x^2-2x+1)$$ What should I do after this? 
 [/INST] Assuming that $(x,y)$ is an integer solution, we have:
$$(2y+1)^2 = 4x^4+4x^3+4x^2+4x+1. \tag{1}$$
The RHS of $(1)$ is greater than $(2x^2+x)^2$ unless $x=-1$. In such a case $y\in\{-1,0\}$ gives two solutions. The RHS of $(1)$ is less than $(2x^2+x+1)^2$ for any $x$ outside $[0,2]$, so we can have integer solutions only if $x\in\{-1,0,1,2\}$. Now computing them is easy:
$$\{(-1,-1),(-1,0),(0,-1),(0,0),(2,-6),(2,5)\}.\tag{2}$$
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