# Solving the Diophantine equation $y^2 = x^4+x+ 2$ for $x,y \in \mathbb{Z}$

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.

If $$x\geq -1$$, then $$x^4 Therefore, for $$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 $$(x^2-2)^2
Therefore, for $$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 $$(-2,\pm4)$$.

• Can you further explain the statement: there exists a solution iff $x^4+x+2=(x^2+1)^2$? – Jens Wagemaker Dec 7 '18 at 20:58
• 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: $(a+1)^2$. – Batominovski Dec 7 '18 at 21:03

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.

• 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.. – Jens Wagemaker Dec 7 '18 at 20:05
• Diophantine equations are not for exams. They are for pleasure! – Dietrich Burde Dec 7 '18 at 20:41
• I wonder what grade I get if I write that on the exam :p – Jens Wagemaker Dec 7 '18 at 20:51