Are there infinitely many solutions (m,n) to $\left\lvert3^m - 2^n\right\rvert$= 1 $\forall$ m,n $\in$ $\mathbb{Z}^+$ ?
For this argument, (which is part of a larger proof I'm working on) it doesn't matter whether $3^m$ > $2^n$ or $3^m$ < $2^n$ so long as m,n are positive and the difference between $3^m$ and $2^n$is 1; (therefore I put absolute values around it); my point is to show (or to know if it's false) that periodically, for infinitely high positive integers m,n, $3^m$ and $2^n$ can differ by a measure of 1 infinitely many times; therefore, equivalently I could ask,
Are there infinitely many solutions (m,n)$\in$$\mathbb{Z}^+$ such that $\frac {3^m}{2^n}$ = z +1 for some positive integer z, OR $\frac {2^n}{3^m}$ = z + 1 for some positive integer z (where 1 is a remainder)?
Two obvious solutions are where m,n=1 and where m=2 and n=3, but are there infinitely many solutions?
My work on the problem has gone in two directions, one algebraic and the other geometric; algebraically I've attempted to reduce the exponentiation operation to a recursive arithmetic operation so that e.g.
$2^3$ = 8
$2^4$ = 8 $+^n$ 8
$2^5$ = 8 + 8 $+^n_2$ 8 +8
$2^6$ = 8 + 8 + 8 + 8 $+^n_3$ 8 + 8 + 8 + 8
...
$2^n$ = 8 + 8 + ... $+^n_r$ 8 + 8 for some r= n+1 number of recursive operations for n>3 , and likewise for
$3^2$ = 9
$3^3$ = 9 $+^m$ (9 + 9)
$3^4$ = 9 + 9 + 9 $+^m_2$ (9 + 9 + 9)
$3^5$ = 9 + 9 + 9 + 9 + 9 + 9 $+^m_3$ (9 + 9 + 9 + 9 + 9 + 9)
...
$3^m$ = 9 + 9 + 9 +... $+^m_q$ 9 + 9 + 9 for some q= m+1 number of recursions for m>2 (as an amateur/non-specialist I don't really know if my notation is correct)
So that the powers products of $3^m$ and $2^n$ for m>2 and n>3 can be expressed as the sums of multiples of 9 and 8 respectively,
$3^m$ = $\sum_{i=3}^q 9i$ for i multiples of 9 and
$2^n$ = $\sum_{i=4}^r 8i$ for i multiples of 8
so that the relation $3^m$R$2^n$ can be re-written in terms of the parity of the multiples of 9 and 8 so that since all $3^m$ are odd and all $2^n$ are even,
$\sum_{i=3}^q 9i$ = 9(2k+1)i + 9(2k+1)i + 9(2k+1)i +... $9(2k+1)_q$(i)
$\sum_{i=4}^r 8i$ = 8(2k)i + 8(2k)i + 8(2k)i + ... $8(2k)_r$(i)
For however many odd multiples q of 9 for 9i , and however many even multiples r of 8 for 8i, $\forall k$>0 $\in$$\mathbb{N}$.
(Please excuse me if I've made so many mistakes) But since I'm trying to prove it true (which maybe I'm completely wrong-headed in thinking it may be), possibly there is some proof by induction perhaps? that for arbitrarily large values of m,n , at some point the difference between some odd and some even will be exactly 1?
Thank you so much for taking the time to help me; (I will spare you the geometric considerations, since for starters I'm new to the site and typesetting geometric figures might prove difficult to me, and also, more importantly my entire enterprise may be either trivial or misguided, since as I said I am an amateur, perhaps just barely more than a dilettante)
