$x^2 + 1 = z = y^3 - 1$

Why $z = 26 $ and only $26$ ? Is there an elementary proof of that ?

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    $\begingroup$ One proof comes down to factoring in $\mathbb{Z}[\sqrt{-2}]$. I have a feeling I have done this on MSE, and it has probably been done by others. $\endgroup$ Commented Jul 21, 2014 at 23:37
  • $\begingroup$ This question seems particularly relevant, along with the MathOverflow question linked there. $\endgroup$ Commented Jul 21, 2014 at 23:57
  • $\begingroup$ Meanwhile, the only way to get the square bigger by exactly 2 is $(-1,1)$ as $(-1)^3 = -1.$ tnt.math.se.tmu.ac.jp/simath/MORDELL/MORDELL+ $\endgroup$
    – Will Jagy
    Commented Jul 22, 2014 at 0:32
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    $\begingroup$ Is there something more that you're confused on op? You haven't accepted any of the answers or commented to clarify. $\endgroup$ Commented Jul 25, 2014 at 4:58
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    $\begingroup$ @Adam Hughes, I appreciate for you effort. Your explanation was great. Thank you for your time. $\endgroup$ Commented Jul 25, 2014 at 6:49

2 Answers 2


Instead write $$x^2+2=y^3$$ so that $x^2+2=(x+\sqrt{-2})(x-\sqrt{-2})$ is a norm from the integer ring $\Bbb Z[\sqrt{-2}]$ which is Euclidean. So then it is clear that $\gcd(x+\sqrt{-2},x-\sqrt{-2})|\sqrt{-2}$, so that if $\sqrt{-2}\not |x$ they are coprime. Since $x$ is an integer, this means their gcd is $\sqrt{-2}$ iff $x$ is even and 1 otherwise.

Case 1: $\sqrt{-2}|x$ whence $x=2m$ so that we have $x^2+2=4m^2+2=2(m^2+1)$ so $2|y$ and $m=2k+1$ necessarily. But then $x^2+2=4m^2+2=4(2k+1)^2+2=4(4k^2+4k+1)+2=16(k^2+k)+6$ which is not a multiple of 8, a contradiction since $8|y^3$. Hence $x$ is odd which puts us in

Case 2: $\gcd(x+\sqrt{-2},x-\sqrt{-2})=1$

Then $x+\sqrt{-2}=(a+b\sqrt{-2})^3$ and here's the kicker:

$$(a+b\sqrt{-2})^3=a^3+3a^2b\sqrt{-2}-6ab^2-2b^3\sqrt{-2} = (a^3-6ab^2)+(3a^2b-2b^3)\sqrt{-2}$$

Now since $3a^2b-2b^3=b(3a^2-2b^2)=1$ it must be that $b=\pm 1$ and similarly $3a^2-2b^2=\pm 1$. Since $b^2=1$ this means $3a^2=2\pm 1$ clearly $3a^2=1$ is impossible, so $a=\pm 1$ gives us our only solutions.

So $x=a^3-6a=a(-5)$ the only positive possibility is $x=5$, which immediately gives $y=3$.

  • $\begingroup$ In case 1, why must $m=2k+1$? $\endgroup$ Commented Jul 21, 2014 at 23:54
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    $\begingroup$ @BeaumontTaz if $m$ is even then $2(m^2+1)\equiv 2\mod 4$, but $y^3\equiv 0\mod 4$. $\endgroup$ Commented Jul 21, 2014 at 23:55

The equation $y^2 = x^3 - 2$ defines an elliptic curve, and it's known that such a curve has only finitely many integer solutions. I don't think there's a simple way of actually finding all of those solutions, though, at least in the general case. There are algorithmic solutions, but nothing that's easily done by hand.

This paper covers a few instances of the Mordell curve $y^2 = x^3 + k$ using elementary methods, and it specifically includes the case $y^2= x^3 - 2$ you mentioned. The technique depends crucially on the ring of integers in $\mathbb{Q}(\sqrt{k})$ being a unique factorization domain, though, which is not the case in general. It is true for $\mathbb{Q}(\sqrt{-2})$, though, so it turns out to be straightforward (though it does involve a bit of computation).

  • $\begingroup$ By Siegel's theorem, there are only finitely many $K$-integral points on an elliptic curve $C/K$. There can be infinitely many $K$-rational points, though. $\endgroup$
    – anomaly
    Commented Jul 22, 2014 at 0:10

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