$x$ 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?
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asked Aug 20, 2016 at 19:22
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1$\begingroup$ i have found the same solution $\endgroup$– Dr. Sonnhard GraubnerCommented Aug 20, 2016 at 19:29
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3$\begingroup$ 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: math.uconn.edu/~kconrad/blurbs/gradnumthy/mordelleqn1.pdf $\endgroup$– user236182Commented Aug 20, 2016 at 19:35
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2$\begingroup$ The given solution solves it in integers instead of just positive integers. All the integer solutions are $(x,y)=(0,-1),(\pm 1, 0),(\pm 3, 2)$. $\endgroup$– user236182Commented Aug 20, 2016 at 19:37
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3$\begingroup$ It essentially used to be known as Catalan's conjecture. $\endgroup$– Sylvain JulienCommented Aug 20, 2016 at 19:59
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1$\begingroup$ @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. $\endgroup$– Rob ArthanCommented Aug 20, 2016 at 21:59
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1 Answer
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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 $(3,2)$.
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$\begingroup$ Poop. Yep, I need to change some signs and conjugates. $\endgroup$ Commented Aug 22, 2016 at 11:08