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Is the conjecture true?

$3^n - 2^m = 1 $ has infinitely many solutions, where n, m are natural numbers.

More generally, $P^n-Q^m=1$ has infinitely many solution for n, m ϵ {1,2,3,…} where P is odd, Q is even, P and Q have no common factors.

$P^n-Q^m=2$ has infinitely many solution for n, m ϵ {1,2,3,…} where P, Q are both odd, P and Q have no common factors.

This is my first post - I joined this community to look into this.

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  • $\begingroup$ Welcome to MSE! Please use the basic tutorial and quick reference guide and also show the work you have done so far. $\endgroup$
    – Jessie
    Feb 7, 2021 at 18:49
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    $\begingroup$ See the Catalan conjecture, which was proven in 2002. $\endgroup$
    – jlammy
    Feb 7, 2021 at 18:49
  • $\begingroup$ Well, what have you done so far? What solutions have you found so far? If there are infinitely many, there is probably some pattern. So find some! $\endgroup$
    – almagest
    Feb 7, 2021 at 18:49
  • $\begingroup$ Thank you. I'd mark the Catalan conjecture as answering my question. But I don't see how. $\endgroup$ Feb 7, 2021 at 18:52
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    $\begingroup$ There is only ONE pair of consecutive perfect powers (if we exclude $0$ and $1$) , namely $8$ and $9$. This has been proven in the mean time, for other differences the problem is open (See Pillai's conjecture) $\endgroup$
    – Peter
    Feb 7, 2021 at 18:56

1 Answer 1

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Mihăilescu's theorem, before 2002 known as Catalan's conjecture, states that:

The only natural numbers $m,n>1$ and $P,Q>0$ satisfying $$P^n-Q^m=1,$$are $(m,n,P,Q)=(2,3,3,2)$.

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