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

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

marked as duplicate by Gerry Myerson, William, Ross Millikan, Rudy the Reindeer, Chris EagleOct 3 '12 at 10:13

• "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? – Steven Stadnicki Sep 12 '12 at 15:34
• Is this not covered by Catalan's conjecture, which is actually a theorem rather than a conjecture? – Old John Sep 12 '12 at 15:51
• 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.... – Gottfried Helms Sep 12 '12 at 16:23
• If it's not too much trouble, I would like to ask for a reference of the fact that $19\mid 2^l + 1 \iff 27\mid 2^l + 1$. – EuYu Sep 12 '12 at 16:54

take your original equation, and pass $2^l$ to the other side of the equal.
$3^k$ and $2^l +1$ are equal if they have the same prime dividers. What are the prime divider of $3^k$? Only 3. What are the prime divider of $27$? Only 3.
So you can use your previous questions to find other possibility : $2^l + 1$ must be a multiple of 19, and must be an odd number. How many solutions of this sub-question can you find ? Once you have them it should be easy to find the corresponding value for $k$.