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I was reading a lot of other questions here which are supposed to be similar, but none of the answers gave me a hint, how I can approach this. I wrote a Python program to solve it, but up to $2^n < 1e20$ there were no solutions. So I guess my question should be, how do I prove that this equation has no integer solutions other than the trivial solution n = m = 0?

Rüdi Jehn
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  • $(m,n)=(0,0)$ is one solution – DatBoi Nov 24 '20 at 15:06
  • you can look at the exponent of primefactor $2$ in the expression $(3^{m+1}-1)$ depending on $m$. Say, this is $e_2$ for some $m$. Then by the lhs $2m$ must have the form $2^{e_2} \cdot x$ Then you can look at the primefactor $3$ in the parenthese on the lhs depending on $n$. This gives a start - well, just to have at least one nontrivial step with it...unfortunately I've not really my brain at this by the moment.. – Gottfried Helms Dec 06 '20 at 12:35
  • @GottfriedHelms: are you suggesting to try $(3^{m+1} - 1) = 2^{e_2}$ ? This is impossible. It would lead to $$2m (2^{n+1} - 1) = n 2^{e_2}$$ or $$m (2^{n+1} - 1) = n 2^a.$$ This means $m$ must be a multiple of $2^a$ because $ (2^{n+1} - 1)$ has no factor of 2. So if $m = k 2^a$ then we try to solve $$k (2^{n+1} - 1) = n $$ or $$k 2^{n+1} = n + k$$. The lhs is $k2^n + k2^n$. The first summand is > n and the second is >k and therefore the lhs > rhs. Finito, no solution. I need to look somewhere else – Rüdi Jehn Dec 07 '20 at 13:58
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    Hmm. Let $m+1=2^A \cdot x$ with $x$ odd. Then the primefactor $2$ must occur on the rhs to the exponent $A+2$ such that we have the rhs $n \cdot 2^{A+2} \cdot y$ At the lhs: because if $m+1$ contains the primefactor $2$ then $m$ must be odd, so we have something like $2 (2^Ax-1)(2^{n+1}-1) = n 4\cdot 2^A y$ which is impossible. Thus $m+1$ must be odd (or: $A=0$ ) . Then the rhs has the primefactor $2$ only to the exponent $1$. But then, after cancelling $2$ in rhs & lhs, $m$ on the lhs is even, but the rhs is odd. Hope no hole in this - I just had such a vague idea, without pen&paper... – Gottfried Helms Dec 07 '20 at 17:02
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    I am still speechless looking at the simplicity of this proof. It took me a good hour to understand the second sentence that the rhs must be a multiple of $2^{A+2}$. Without pen&paper?? You must have a real deep relationship with numbers. Viele Grüße von Darmstadt nach Kassel! – Rüdi Jehn Dec 07 '20 at 19:43
  • :-) Thanks for the very nice compliment! And Grüsse back from Kassel nach Darmstadt! It's a time ago I experimented with the question of primefactor-decomposition of cyclotomic forms ($a^n-b^n % p^k$) also in the view of "Fermat-quotients". A very nice essay (though amateurish) you might like: http://go.helms-net.de/math/expdioph/CyclicSubgroups_work.pdf and some of my other fiddlings in the index http://go.helms-net.de/math/. (and about Fermatquotients http://go.helms-net.de/math/expdioph/fermatquotients.pdf ) Have fun! – Gottfried Helms Dec 08 '20 at 20:57
  • Dear @GottfriedHelms, I had a look at your treatize "Exponential diophantine problems: The power of cyclic subgroups" and on bottom of page 5 in Example 3, there is ${b^n -1, p} \ge 1+ {n, p}$ if ... In our case $b=3$, $n=2^Ax$ and $p=2$. This will give ${b^n -1, p} \ge 1+A$, but you said it is $2+A$ (which I verified by writing out the factors of $3^{2^Ax} -1$ and counting the factors of 2). But I guess you were not counting, instead you have a very short proof that ${3^{2^Ax} -1, 2} = 2+A$? – Rüdi Jehn Dec 10 '20 at 16:59
  • Hi Rüdi - this type of "jump" from $ {3^1-1,2} = 1$ to $ {3^2-1,2} = 3$ is unique to the primefactor $2$ . It can be explained by writing out the formula $ (3^2-1)=(3^1-1) \cdot (3^1+1)= 2^1 \cdot 2^2 = 2^3$ and the difference between $3-1$ and $3+1$ is just divisible by that primefactor $2$, which cannot occur with any other primefactor . I've tried to give it a "good" formal notation with the $\beta()$-function vaguely mentioned at page $9$, although (perhaps unfortunately) I did not give it much prominence... :( – Gottfried Helms Dec 10 '20 at 17:28
  • One might say: the $\alpha()$-function gives the exponent of a primefactor $p$ at its first occurence in $b^n-1$, the $\beta()$-function at its first occurence in $b^n+1$. But this is only significant for the primefactor $p=2$ because it can occur in both functions at the same exponent, "simultaneously". This part of my formalism might be a bit weak, and be improvable - I just had no idea how to make it smoother. – Gottfried Helms Dec 10 '20 at 18:19
  • WolframAlpha offers some insights here and here. – poetasis Jan 07 '21 at 18:48

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