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Find all integer solutions for $$x^3+1=y^2.$$

Attempt: By guessing, I found five pairs of integer solutions for the equation: $(2, \pm 3)$, $(0, 1)$, $(-1, 0)$ and $(0, -1)$, but really I don't know how to solve it analytically without guessing. Some people lead this problem to the Catalan?

Asaf Karagila
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Shane Dizzy Sukardy
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3 Answers3

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Hint: This an example of "Mordell's Equation" - curves of the form $y^2 = x^3 + D$ (in your case D = 1). Many things are known about its integral solutions. You might find this article useful , you can also proceed using this answer and also this one

Addendum: Mordell spent many years of his life studying integral solutions of the equation $y^2 = x^3 + k$, where $k$ is a fixed nonzero integer. The equation could be justified as having interest because it's one of the simplest examples of an elliptic curve, but it's important for a better reason. The $abc$-conjecture, which has connections to many other problems, does not at first look like it is about Mordell's equation. However, the $abc$ conjecture turns out to be equivalent to specific upper bounds on relatively prime integral solutions $(x,y)$ to Mordell's equation $y^2 = x^3 + k$ in terms of the parameter $k$. So, as Barry Mazur once remarked, the Mordell equation is a far more central topic to all of number theory than its rather special appearance suggests.

zeraoulia rafik
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    This is not a hint, this is a reference. – Wojowu Jun 02 '20 at 22:56
  • The titled question should be follow the linked answer we do not need to repeat the same steps which are montioned in the linked answer – zeraoulia rafik Jun 02 '20 at 22:57
  • I am not complaining about the answer per se (though I could, this is something I would see more fit as a comment). I am only complaining about the use of the word "Hint" to refer to something that's a link to a full solution and not a hint. – Wojowu Jun 02 '20 at 23:06
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Partial answer.

The elementary approach is to write

$$x^3=y^2-1=(y-1)(y+1)$$

Now, when $y$ is even then $y-1$ and $y+1$ are relatively prime, so $y-1=w^3$ and $y+1=z^3$ for integers $w,z.$ But $z^3-w^3=2$ is only possible if $(w,z)=(-1,1),$ or $y=0, x=-1.$

The case $y$ odd is a bit harder, but not too hard. It becomes equivalent to solving the equation: $$2n^3=m(m+1)$$ where $n=x/2$ and $m=(y-1)/2.$

Use that $m$ and $m+1$ are relatively prime.

This reduces to the case $$u^3-2v^3=\pm1.\tag{1}$$ That’s a harder equation to solve, perhaps.

Once you have $u,v$, you get $y=u^3+2v^3$ and $x=2uv.$

The solutions to (1) are $(u,v)=(1,1),(-1,-1),(1,0),(-1,0)$ yield $(x,y)=(2,3),(2,-3),(0,1),$ and $(0,-1)$ respectively.

That there are no other solutions seems non-trivial. Perhaps you can rewrite it as:

$$(x-y)(x^2+xy+y^2)=x^3-y^3=y^3\pm 1=(y\pm 1)(y^2\mp y+1)$$

but I'm not seeing an obvious approach from there.

Thomas Andrews
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According to Sage, all integral points are below.

E=EllipticCurve([0,0,0,0,1])
E.integral_points()

$(0, \pm 1),(2, \pm 3),(-1,0)$

Tomita
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