Exercise 5.4.10 (Equation $x^4 + 2x^3 + 2x^2 + 2x+5 = y^2$)

Proof. $\text{∙}$ We give first the solutions such that $|x|\le 2$.

If $x=2$, then $y=\pm 7$, if $x=-2$, then $y=\pm 3$ and if $x=-1$ then $y=\pm 2$. If $x=1$ then $y^2=12$ has no solution, and if $x=0$, $y^2=5$ has no solutions. There are exactly $6$ solutions such that $|x|\le 2$, which are

$\text{∙}$ We prove that the equation $x^4+2x^3+2x^2+2x+5=y^2$ has no solution $(x,y)\in {\mathbb{Z}}^2$ such that $|x|>2$.

Suppose for the sake of contradiction that $x^4+2x^3+2x^2+2x+5=y^2$ and $|x|>2$. Then $-x^2+4<0$, so $y^2={(x^2+x+1)}^2-x^2+4<{(x^2+x+1)}^2$.

Moreover $y^2-{(x^2+x)}^2=x^2+2x+5={(x+1)}^2+4>0$. Therefore

Here $|x|>2$, a fortiori $x\notin [-1,0]$, thus $x^2+x>0$, and $x^2+x+1$ is always positive. Then (1) implies

so the integer $n=|y|-(x^2+x)$ satisfies $0<n<1$. This is impossible. Therefore the equation $x^4+2x^3+2x^2+2x+5=y^2$ has no solution $(x,y)\in {\mathbb{Z}}^2$ such that $|x|>2$.

In conclusion, the integral solutions of the equation $x^4+2x^3+2x^2+2x+5=y^2$ are $(-2,3),(-2,-3),(-1,2),(-1,-2),(2,7)(2,-7).$ □