# 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

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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

"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

## 1 Answer

If by "iff" you mean "if and only if" (just to be sure, I didn't learn maths with english) then your problem is almost solved :

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 $k$.

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