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Find integers $x$ and $y$ such that $$x^3+x-y^2=1.$$

My try:

$$x^3+x-y^2=1 \implies x^3+x-1=y^2.$$

Now, when $x^3+x-1$ is a perfect square?

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  • $\begingroup$ hint: left hand side is even number , so $y^2+1$ muse be even $\to y $ must be odd number . $\endgroup$ – Khosrotash Jan 21 '17 at 14:05
  • $\begingroup$ If I'm not mistaken, using the unique factorisation over the ring $\mathbb{Z}[\text{i}]$ of Gaussian integers, this question is equivalent to finding $u,\alpha,\beta\in\mathbb{Z}[\text{i}]$ such that $\text{Im}(u\alpha\beta)=1$ and $\alpha\bar{\beta}-u\bar{u}=\text{i}$. (That is, $x=u\bar{u}$ and $y=u\alpha\beta-\text{i}$.) I expect an infinite family of solutions. $\endgroup$ – Batominovski Jan 21 '17 at 14:20
  • $\begingroup$ Also $x, y$ has to be mutual prime $\endgroup$ – user261263 Jan 21 '17 at 14:54
  • $\begingroup$ I find $(1,1), (2,3), (13,47)$ as small solutions $\endgroup$ – Ross Millikan Jan 21 '17 at 15:26
  • $\begingroup$ I've wrote a computer program to find solutions. For $x \lt 10000000000$ the solutions are: (1,1) (2, 3) (13, 47), (5112308218, 1226294255). I think @Batominovski is right, there might be an infinite family of solutions. $\endgroup$ – user261263 Jan 21 '17 at 15:28
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The elliptic curve $y^2=x^3+x-1$ has only finitely many integral points, according to the magma online calculator - see here at MO, namely the points $$ (x,y)=(1,\pm 1),(2,\pm3),(13,\pm 47). $$

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