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Possible Duplicate:
$|2^x-3^y|=1$ has only three natural pairs as solutions

As the title says:

Find all positive integers k and l such that $3^k-2^l=1$

In previous questions it is proved that $2^l+1$ is divisible by $27$ iff it is divisible by $19$, though I'm not sure how that is meant to help.

I've brute forced it, and got that $k = l = 1$ works, as does $k = 2$ and $l = 3$, but I can't figure out how to prove that these are the only solutions.

Any help would be much appreciated

  • 2
    "In previous questions it is proved that $2^l+1$ is divisible by 27 iff it is divisible by 19" - if $2^l+1$ were of the form $3^k$ for $k\geq 3$, what would that then say about its divisibility? What primes can a number of the form $3^k$ be divisible by?2012-09-12
  • 0
    Is this not covered by [Catalan's conjecture](http://en.wikipedia.org/wiki/Catalan's_conjecture), which is actually a theorem rather than a conjecture?2012-09-12
  • 0
    you could rearrange to $3^k-1 = 2^l$ and look, whether you can find out a rule, how the powers of $2$ occur in the lhs, dependent on the change of $k$. Fermat and Euler shall help....2012-09-12
  • 0
    If it's not too much trouble, I would like to ask for a reference of the fact that $19\mid 2^l + 1 \iff 27\mid 2^l + 1$.2012-09-12

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