Is $2^m + 1$ ever a power of 3, for $m > 3$?

Question

Let $m,n \in \mathbb{Z}$. For $m > 3$, can $2^m + 1 = 3^n$?

Answer 1

If $m\gt1$, then $2^m+1\equiv1$ mod $4$, which implies $2^m+1$ can only be an even power of $3$, say $3^{2n}$. But then $2^m=3^{2n}-1=(3^n+1)(3^n-1)$, which implies $3^n+1$ and $3^n-1$ are both powers of $2$. This is only the case for $n=1$.

Remarks: Will Jagy's answer is well worth reading, because his approach applies in a great many settings where the solution cannot be obtained as simply as here. And regarding Ross Millian's mention of Catalan, it's always good to know when there's a sledgehammer available....

Answer 2

I did a number of these by a consistent method in September and October 2016, some links below. I've always liked that $343 = 243 + 100;$ one of the links shows that is the largest solution to $7^u - 3^v = 100. $ I had not done this one, apparently. Below is the investigation by elementary means. Give me some time, i will typeset the outcome.

The main ingredient used is the multiplicative order, see property (1) in this jpeg:

We have $$ 3^u = 2^v + 1. $$ Subtract $9$ from both sides, we have $$ 3^u - 9 = 2^v - 8. $$ Taking $u = x + 2$ and $v = 3 + y$ gives $9 \cdot 3^x - 9 = 8 \cdot 2^y - 8,$ or $$ 9 (3^x - 1 ) = 8 (2^y - 1). $$ We think this is only possible with $x=y=0,$ so we assume there is a solution with $x,y > 0$ and get a contradiction.

First we have $9 |(2^y - 1),$ or $$ 2^y \equiv 1 \pmod 9.$$ This tells us that $ 6 | y. $ Meanwhile $$ 2^6 - 1 = 63 = 3^2 \cdot 7. $$ Therefore $7 | (3^x - 1).$ In turn, we find $6 | x.$

Next $$ 3^6 - 1 = 728 = 2^3 \cdot 7 \cdot 13. $$ Therefore $13 | (2^y - 1).$ In turn, we find $12 | y.$

Next $$ 2^{12} - 1 = 4095 = 3^2 \cdot 5 \cdot 7 \cdot 13. $$ Therefore $5 | (3^x - 1).$ In turn, we find $4 | x.$

Finally, $3^x - 1$ is divisible by $3^4 - 1 = 80.$ In particular, $3^x - 1$ is divisible by $2^4 = 16.$ However, this contradicts $2^y - 1 \neq 0$ and $y > 0.$ $$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$

This method does well when the numbers are not too large.

Exponential Diophantine equation $7^y + 2 = 3^x$

Elementary solution of exponential Diophantine equation $2^x - 3^y = 7$.

Elementary solution of exponential Diophantine equation $2^x - 3^y = 7$.

Finding solutions to the diophantine equation $7^a=3^b+100$

9 ( 3^x - 1) = 8 ( 2^y - 1)
ASSUME x,y > 0.

jagy@phobeusjunior:~$ ./order 3 8
8     2 = 2
jagy@phobeusjunior:~$ ./order 2 9
9     6 = 2 * 3
jagy@phobeusjunior:~$ 
jagy@phobeusjunior:~$ ./order 3 16
16     4 = 2^2
jagy@phobeusjunior:~$ ./order 2 27
27    18 = 2 * 3^2
jagy@phobeusjunior:~$ 

Given: 2 | x  ,    6 | y

WANT  4 | x  ,   9 | y
jagy@phobeusjunior:~$ ./prime_power_minus_one   2  6
    2^6 - 1    = 3^2  7
jagy@phobeusjunior:~$ ./order 3 7
7     6 = 2 * 3
=====================================
THEREFORE  6 | x  
jagy@phobeusjunior:~$ ./prime_power_minus_one   3  6
    3^6 - 1    = 2^3 7  13
jagy@phobeusjunior:~$ ./order 2 13
13    12 = 2^2 * 3
==============================
Therefore 12 | y
jagy@phobeusjunior:~$ ./prime_power_minus_one   2  12
    2^12 - 1    = 3^2 5 7  13
jagy@phobeusjunior:~$ ./order 3  5
5     4 = 2^2
========================
Therefore 4 | x  

Therefore 16 | (3^x - 1). This contradicts 
9 ( 3^x - 1) = 8 ( 2^y - 1)  with x,y > 0.

Answer 3

No. Catalan's conjecture (now Mihăilescu's theorem) says that the only perfect powers that differ by $1$ are $3^2=9$ and $2^3=8$ I am sure there is a much simpler proof for this specific case.