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Are there any positive integer solutions to $2^n-3^m=1$ with $n,m>2$ ?

By way of justifying the question, I've found lots of info on what happens where $m=n$ (mostly FLT variantions, Darmon + Merel,...) but don't really know where to look for $m\not=n$.

Also it's pretty obvious that you can't have solutions to similar equations, e.g. $2^n-3^m=2$. There are no solutions for $n,m<1000$ asside from $n=2,m=3$. It seems pretty likely to me that it should happen for some large numbers some point though.

Are there any theorms I don't know about regarding primes $p,q$ and $p^n-q^m=k$, $k \in \mathbb{N}$ that might rule out a solution or help me find one?

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Didn't Gersonides do this in 1343? en.wikipedia.org/wiki/Gersonides "One year later, at the request of the bishop of Meaux, he wrote The Harmony of Numbers in which he considers a problem of Philippe de Vitry involving so-called harmonic numbers, which have the form $2^m\cdot 3^n$. The problem was to characterize all pairs of harmonic numbers differing by 1. Gersonides proved that there are only four such pairs: (1,2), (2,3), (3,4) and (8,9)." Ivars Peterson gives an easy proof at maa.org/mathland/mathtrek_1_25_99.html –  Junkie Jul 1 '11 at 12:25
Thanks guys, that was much more interesting than I expected! –  Kevin Jul 1 '11 at 13:22
Hmm, the question is already answered, and an answer is also accepted, so this is just an addendum. I'd simply reorder the equation into $2^n-1=3^m$ and use Euler's phi-function for the primefactors of the lhs and the powers of 3: to have a power k of 3 as factor of $2^n-1$ n must have the form $x*\varphi(3^k)=x*2*3^{k-1}$ where x is coprime to 2 and 3. After that, the lhs has additional (prime-)factors due to the $\varphi$-function for nontrivial n>1 except if n=6; here we can use the szigmondy-theorem or a simple comparision of the growthrate of the lhs and rhs, if k>2. –  Gottfried Helms Jul 1 '11 at 18:56

4 Answers 4

up vote 33 down vote accepted

Here is the proof of Gersonides [Levi ben Gershon] (1343) for $2^n-3^m=1$. It uses nothing more that arithmetic modulo $8$.

Case I: $m$ is even. Then $3^m$ is 1 mod 4, so $2^n$ is 2 mod 4, implying $n=1$ and $m=0$.

Case II: $m$ is odd. Then $3^m$ is 3 mod 8, so $2^n$ is 4 mod 8, implying $n=2$ and $m=1$.

The alternative equation $3^m-2^n=1$ follows similarly when $m$ is odd, but is a bit more tricky when $m$ is even (hint, factor $2^n=3^m-1=(3^{m/2}+1)(3^{m/2}-1)$ and argue from there).

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Yes, indeed, the argument of Levi ben Gershon didn't change during the last year: mathoverflow.net/questions/29926/… –  Wadim Zudilin Jul 2 '11 at 17:30

The equation $p^n-q^m=k$ is a special case of the $S$-unit equation, so it has finitely many solutions by a theorem of Siegel. Using linear forms in logarithms you can get an explicit upper bound for the size of the solutions and, in principle, find all of them. In practice, the bounds may be too big in general, but I am sure $2^n-3^m=1$ has been done (it also follows from Catalan, as Todd has just commented). Check out Baker's book on Transcendence.

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It is the content of Catalan conjecture That is to say, that the only solution in the natural numbers of $x^a -y^b= 1$ for $ x,a, y,b > 1$ is $x=3, a=2, y=2, b=3$

it was proved in 2002 by Preda Mihăilescu.


a good refernce for a self contained proof is the book by Rene' Schoof: "Catalan's Conjecture", Universitext, Springer-Verlag, 2008.

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The question is much more special than Catalan's conjecture: the numbers x and y are already specialized to 3 and 2. So it is misleading to say the question "is the content" of that harder problem. Still, given that Catalan's conjecture is solved, making a link between them is worthwhile for perspective. –  KConrad Jul 1 '11 at 17:31
Of course you are right, I was a bit hasty –  Valerio Talamanca Jul 3 '11 at 12:24

There is another method that allows one to handle $a^{n}-b^{m}=k$ (here I call the bases $a$ and $b$ because primality is not important to how the method works). Specifically, if one has a solution, it allows a larger solution to be found, or proven to not exist.

As an example of this method, it is easy to outline a proof that $2^{n} - 5^{m} = 3$ has no solutions larger than $(m,n)=(3,7)$:

Suppose $m>3$, $n>7$, and $2^{n}-5^{m}=3$.
Rewrite the equation as $2^{n}=5^{m}+3$.
Now, to use the largest solution we know, subtract $2^{7}=5^{3}+3$ from both sides to obtain $2^{7}(2^{n-7}-1)=5^{3}(5^{m-3}-1)$.
Since $m$ and $n$ give a solution larger than the one we know, both sides are positive integers. Since $n>7$, the highest power of $2$ dividing the right side is $2^{7}$.

Since the order of $5$ in $(\mathbb{Z}/(128))^{\times}$ is $32$, we know that $32$ divides $m-3$, and $5^{32}-1$ divides both sides. Then $29423041$, as a prime factor of $5^{32}-1$, divides both sides. Then $29423041$ divides $2^{n-7}-1$, so since the order of $2$ in $(\mathbb{Z}/(29423041))^{\times}$ is $122596$, $2^{122596}-1$ divides both sides. (This is probably not a profitable direction to take, but it can work as an illustration of the method.)

The contradiction would be obtained by concluding that $5^{4}$ divides the right side, or $2^{8}$ divides the left side.
In the case where a larger solution exists, the ability to bounce back and forth between the two sides of the equation only goes as far as concluding that the larger solution (that is, the common value of both sides when the larger solution is plugged in) minus the common value from the known solution divides both sides.

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