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$x$ and $y$ are positive integers greater than $1$, such that $x^2-y^3=1$. what are the possible values of $x$ and $y$?

Question: I find, imputing values of $x$ and $y$, $x=3$ and $y=2$ is a solution.

Is there any other solution of this equation?

user236182
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Mahmudul Hasan
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    i have found the same solution – Dr. Sonnhard Graubner Aug 20 '16 at 19:29
  • what was your approach? any equation solving? – Mahmudul Hasan Aug 20 '16 at 19:30
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    This is a Mordell's equation with an elementary (only uses things like unique factorization in $\mathbb Z$) but slightly long (one page) solution. See the pages $7-8$ of this paper: http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/mordelleqn1.pdf – user236182 Aug 20 '16 at 19:35
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    The given solution solves it in integers instead of just positive integers. All the integer solutions are $(x,y)=(0,-1),(\pm 1, 0),(\pm 3, 2)$. – user236182 Aug 20 '16 at 19:37
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    It essentially used to be known as Catalan's conjecture. – Sylvain Julien Aug 20 '16 at 19:59
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    @user236182: irritatingly, the nice paper that you cite reduces this particular Mordell problem to solutions of the equation $a^3 - 2b^3 = 1$ but does not give any further details. So it is not clear how to complete the proof from that reference. – Rob Arthan Aug 20 '16 at 21:59
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    @RobArthan Oops, you're right, I'm sorry. The proof is not elementary and I was wrong in one of the comments. The fact that $a^3-2b^3=1$ has only two integral solutions $(a,b)=(1,0),(-1,-1)$ follows from a theorem by Skolem, namely: For any integer $d \ne 0$, there exists at most one pair $(x,y) \in \mathbb{Z}^2$ with $y \ne 0$ such that $x^3 + dy^3 = 1$. See http://math.stackexchange.com/a/1267684/236182. – user236182 Aug 20 '16 at 23:11
  • @user236182: thanks for the extra information. By coincidence, I too was thinking about the $x^2-y^3=1$ question recently. I had worked out the reduction to $a^2-2b^3=1$ for myself and then found the paper by Conrad, only to be disappointed to find that it didn't complete the proof. I am delighted to learn that the problem was considered and solved by Skolem, who is one of my great heroes. – Rob Arthan Aug 20 '16 at 23:23

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If $x$ is even, write $y^3=x^2-1 = (x-1)(x+1)$. The last two factors are odd and differ by 2, so they are coprime. From the equation, they are both cubes. So we have a pair of cubes which differ by 2, contradiction. So $x$ is odd.

Then write $x^2 = y^3+1 = (y+1)(y+w)(y+w^2)$, where $w$ is a primitive cube root of unity. We can emulate the first paragraph, after some work. We need to know that $\mathbb{Z}[w]$ is a UFD. We need to show $y+w^a$ and $y+w^b$, with $a\neq b$, are relatively prime in this ring. Then each $y+w^a$ is a perfect square. Then we note that squares (and associates thereof) of the algebraic integers in $\mathbb{Z}[w]$ are far apart, while the terms $y+1$, $y+w$, and $y+w^2$ are close to each other. This eliminates all but a small set of possibilities, which are easy to eliminate, leaving only $(1,0)$ and $(3,2)$.

B. Goddard
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