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I know it's a special case of catalan's conjecture.Wiki says its only solution is $(3,2)$.But I cannot work it out.Any help will be appreciated.

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    $\begingroup$ @Tree23- I feel like proving this is much easier than proving the Catalan conjecture. You just have to do trivial analysis, and not worry about the kind of math involved in proving the Catalan conjecture $\endgroup$ – Anju George May 12 at 13:51
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You can re-write the expression as $y^3=x^2-1=(x+1)(x-1)$. Now we can do some analysis. The only common factor that $x+1$ and $x-1$ can have is $2$.

  1. Case 1- They have no common factor. In that case, both $x+1$ and $x-1$ are cubes. This is impossible, as the smallest difference between cubes is $2^3-1^3=7$.
  2. Case 2- They have one common factor- namely $2$. Then for $(x-1)(x+1)$ to be a cube, $x-1=2a^3$ and $x+1=2^2b^3$, or $x-1=2^2a^3$ and $x+1=2b^3$. Here $a,b$ are positive numbers different from $2$. Hence, $|2a^3-2^2b^3|=2$ or $|2^2a^3-2b^3|=2$. We can now see that the only way that this is possible is that $a=b=1$.

Hence, $x=3, y=2$ is the only solution

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  • $\begingroup$ In case 2 won't one of the two factors be $4$ times a cube? Otherwise their product is not a cube. $\endgroup$ – Jyrki Lahtonen May 12 at 14:01
  • $\begingroup$ @JyrkiLahtonen- Yes I have now edited the answer. $\endgroup$ – Anju George May 12 at 14:14
  • $\begingroup$ @HagenvonEitzen- both $x,y>1$. I have also now edited the answer $\endgroup$ – Anju George May 12 at 14:15
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    $\begingroup$ And how can we see that e.g. $|a^3-2b^3|=1$ is only possible with $a=b=1$? $\endgroup$ – Hagen von Eitzen May 12 at 15:03
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Are you looking for some specific solution? Do you mean $3^2-2^3=9-8=1$?

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  • $\begingroup$ Sorry I just forgot to say that Wiki said its only solution with$x,y>1$ is$(3,2)$.I don't know how to prove it. $\endgroup$ – Tree23 May 12 at 13:21
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Not finished!

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

Case 1. $x$ is even then $x+1$ and $x-1$ are relatively prime so $$ x-1 = a^3$$ $$x+1 = b^3$$ where $a<b$ are relatively prime and $ab=y$. Now we have $$2=(b-a)(b^2+ab+a^2)$$ and so:

  • $b-a=1$ and $b^2+ab+a^2 =2$ so $$ 3a^2+3a+1 =2$$ and this is impossibile.

  • $b-a=2$ and $b^2+ab+a^2 =1$ so $$ 3a^2+6a+4=1$$ so $a=-1$ and $b=1$ so $y=-2$ and $x=0$ whivh is not ok again.

Case 2. $x=2z+1$ then $y=2t$ so $$z(z+1)=2t^3$$ Since $z,z+1$ are relatively prime we have

  • $z=2a^3$ and $z+1=b^3$, so $2a^3=b^3-1= (b-1)(b^2+b+1)$. So $$b-1 = 2p^3$$ $$b^2+b+1=q^3$$ where $p,q$ are relatively prime...

  • $z=a^3$ and $z+1=2b^3$, so $2b^3= a^3+1=(a+1)(a^2-a+1)$. So $$a+1 = 2p^3$$ $$a^2-a+1=q^3$$ where $p,q$ are relatively prime...

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