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Find all ordered pairs of integers $(x, y)$ that satisfy $$y^3 = x^6 + 6x^3 + 13.$$

I've found the solutions $(-1, 2)$ and $(2, 5)$. I believe that these are all the integer solutions, but I don't know how to prove it. Could someone please help?

Anon
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    We can rewrite the equation as $$y^3=(x^3+3)^2+4$$ – Peter Oct 12 '17 at 21:59
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    The equation $y^3=z^2+4$ only has the solutions $(2,\pm2)$ and $(5\pm11)$ hence there are no additional integer solutions (Google "Mordell-equation" and "generalized catalan conjecture" for more details) – Peter Oct 12 '17 at 22:07
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    @peter sorry i misread the question I thought it was y on the left side...Sorry – user577215664 Oct 12 '17 at 22:21
  • What's the proof that those are the only solutions? Mordell's theorem only tells me that there are a finite number of solutions, not what those solutions are. – Anon Oct 12 '17 at 22:56

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As Peter said in the comments, this is a special case of the Mordell equation $y^3=z^2+4$ but we can use the fact that $y$ is very nearly $x^2$ to obtain an elementary proof. $$\frac y{x^2}=\left(1+\frac 6{x^3}+\frac{13}{x^6}\right)^{1/3}$$ $$1-\frac 3{|x|^3}<\frac y{x^2}<1+\frac 3{|x|^3}$$ $$|y-x^2|<\frac 3{|x|}$$

And since $y-x^2$ is an integer, it is either 0 which implies $6x^3+13=0$ which is impossible or $\frac 3{|x|}>1$ which wields 5 cases that can be tested by hand.

Sophie
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