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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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    hint: left hand side is even number , so $y^2+1$ muse be even $\to y $ must be odd number . – 2017-01-21
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    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. – 2017-01-21
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    Also $x, y$ has to be mutual prime – 2017-01-21
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    I find $(1,1), (2,3), (13,47)$ as small solutions – 2017-01-21
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    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. – 2017-01-21
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    @EugenCovaci I obtain $5112308218^3+5112308218-1=133613729946263529798154844449 $, but $1226294255^2=1503797599846005025$, so there must be a mistake. – 2017-01-21
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    See here: http://math.stackexchange.com/a/703220/133781 – 2017-01-21

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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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    This might be useful too: http://math.stackexchange.com/questions/32847/integral-points-on-an-elliptic-curve – 2017-01-21
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    @Xam Yes, indeed! My comment [here](http://math.stackexchange.com/questions/703137/cubic-diophantine-equation) points to the same link. – 2017-01-21