Integer solutions of $x^y-z^3=2$

Question

Is it an open question to solve $x^y-z^3=2$ in integers (both positive, zero and negative)? If not, what kind of methods the solution requires?

It is one of the solutions. But I was wondering if we have a proof that some set of triples gives all solutions. — guest, Jun 11 '14 at 17:26
There seem to be several solutions, for example $(29,1,3)$, $(127,1,5)$, $(24391,1,29)$, $(250049,1,63)$, $(328511,1,69)$. — mrf, Jun 11 '14 at 17:27
Yes, yes, yes, there are infinitely many solutions of the form $(z^3+2,1,z)$. The question is, are there other solutions? — Thomas Andrews, Jun 11 '14 at 17:28
Never said it was clever. Those I listed have $x$ as a prime though, making them a little more interesting :) — mrf, Jun 11 '14 at 17:29
@ThomasAndrews, in answer to your question to mrf, yes: $(x,y,z)=(1,3,-1)$. — Barry Cipra, Jun 11 '14 at 17:49
@BarryCipra I don't think solutions of the form $(1, y, -1)$ are any more interesting. — MT_, Jun 11 '14 at 18:04

Will Jagy · Answer 1 · 2014-06-11T18:10:05.313

If $y$ is even, the only solutions to $-t^3 + u^2 = 2$ are $(-1,\pm 1)$

E_+00002: r = 1   t = 1   #III =  1
          E(Q) = <(-1, 1)>
          R =   0.7545769032
           2 integral points
            1. (-1, 1) = 1 * (-1, 1)
            2. (-1, -1) = -(-1, 1)

http://tnt.math.se.tmu.ac.jp/simath/MORDELL/MORDELL+

Things do not change very much if you allow $-2,$ all you get is $27 - 25 = 2.$

E_-00002: r = 1   t = 1   #III =  1
          E(Q) = <(3, 5)>
          R =   1.3495768357
           2 integral points
            1. (3, 5) = 1 * (3, 5)
            2. (3, -5) = -(3, 5)

http://tnt.math.se.tmu.ac.jp/simath/MORDELL/MORDELL-

Also, $y$ cannot be divisible by $3,$ as you can factor the difference of cubes and get a finite number of things to rule out. So, you have $y=5,7,11,13,\ldots$

My guess is that there is a book somewhere on Catalan's conjecture, now proved, with a chapter on changing the difference between powers from $\pm 1$ to $\pm 2, \pm 3,$ and a few other small numbers.

Whatever the exact contents, there is a book called Catalan's Conjecture by Rene Schoof, http://oskicat.berkeley.edu/record=b16478355~S1

Instead of "$y$ cannot be even," don't you mean "If $y$ is even, then $x=\pm1$ and $z=-1$"? — Barry Cipra, Jun 11 '14 at 17:56
@BarryCipra, I see, you think he is allowing all integers, not just positive. Alright. — Will Jagy, Jun 11 '14 at 17:58

Integer solutions of $x^y-z^3=2$

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