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I want to solve the Diophantine equation $y^2 = x^4+x+ 2$ for $x,y \in \mathbb{Z}$.

I already found 4 solutions: $(x,y) = (1,\pm2)$ and $(x,y)=(-2,\pm4)$. It can probably be solved using some factorization argument, but I don't know how.

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If $x\geq -1$, then $$x^4<y^2<(x^2+2)^2\,.$$ Therefore, for $x\geq -1$, there exists a solution iff $x^4+x+2=(x^2+1)^2$. The only integer root of the last equation is $x=1$, yielding the solutions $(x,y)=(1,\pm2)$.

If $x\leq -3$, then $$(x^2-2)^2<y^2<x^4\,.$$
Therefore, for $x\leq -3$, there exists a solution iff $x^4+x+2=(x^2-1)^2$. The last equation does not have an integer root, whence there does not exist a solution in this case.

If $x=-2$, then we have the solutions $(x,y)=(-2,\pm 4)$. Therefore, there are only four solutions $(x,y)\in\mathbb{Z}\times\mathbb{Z}$ to $y^2=x^4+x+2$, which are $(1,\pm2)$ and $(-2,\pm4)$.

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  • $\begingroup$ Can you further explain the statement: there exists a solution iff $x^4+x+2=(x^2+1)^2$? $\endgroup$ – Jens Wagemaker Dec 7 '18 at 20:58
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    $\begingroup$ You have a square $y^2$ lying strictly between two squares $a^2$ and $(a+2)^2$ (here, $a:=x^2$), then surely, $y^2$ is the square in the middle: $(a+1)^2$. $\endgroup$ – Batominovski Dec 7 '18 at 21:03
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There is a "simple method" to solve the Diophantine equation $$ Y^2=X^4+aX^3+bX^2+cX+d $$ in general, see here. The second section does it for the example of $Y^2=X^4-8X^2+8X+1$, but $Y^2=X^4+X+2$ should be even easier.

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  • $\begingroup$ Thanks for the reference. Would you know how to find the bounds on the solutions, without using this theorem? Just for this specific case? Because this theorem is a bit of an overkill, and not something I could remember during an exam.. $\endgroup$ – Jens Wagemaker Dec 7 '18 at 20:05
  • $\begingroup$ Diophantine equations are not for exams. They are for pleasure! $\endgroup$ – Dietrich Burde Dec 7 '18 at 20:41
  • $\begingroup$ I wonder what grade I get if I write that on the exam :p $\endgroup$ – Jens Wagemaker Dec 7 '18 at 20:51

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