Does 2^m = 3^n + r have finitely many solutions for every r? - MathOverflow

Does 2^m = 3^n + r have finitely many solutions for every r?

2010-03-20T02:00:19Z

Is it true that for every integer $r$, the equation $2^m = 3^n + r$ has at most a finite number of integer solutions? I understand that this is a special case of Pillai's conjecture, which is unsolved.

If the statement is true, then can we verify the finiteness of the solution set using modular arithmetic? To be precise, is the following proposition true?

$$\forall r,\ \exists M,\ \exists N,\ \forall m,n \ge N,\ \ 2^m \not\equiv 3^n + r \pmod{M}$$

I have verified the proposition for $0 \le r \le 12$, and found the least possible modulus $M(r)$ for each $r$ in this interval. Note that $M(r) = 2$ if $r$ is even.

$M(1) = 8$, $M(3) = 3$, $M(5) = 1088$, $M(7) = 1632$, $M(9) = 3$, $M(11) = 8$.

Answer by Jonas Meyer for Does 2^m = 3^n + r have finitely many solutions for every r?

2010-03-20T02:41:12Z

I have no comment on your methods, and I know very little about this, but that case of Pillai's conjecture appears to have been solved in the 80's by Stroeker and Tijdeman [Edit: see below]. Here's a paper by Bennett from 2001 that shows more: http://www.math.ubc.ca/~bennett/B-CJM-Pillai.pdf. In particular, the number of solutions is at most 2 for each fixed $r$. More generally, Bennett shows that for fixed integers $a\geq2$, $b\geq2$, and $r\neq0$, there are at most 2 solutions $(m,n)$ to the equation $a^m=b^n+r$. The more general form of Pillai's conjecture allows $a$ and $b$ to vary and appears to still be unsolved.

Edit: What Stroeker and Tijdeman actually did was sharpen the result by showing that except when $r$ is in $\{-1,5,13\}$ , your equation has at most one solution, and that in the exceptional cases it has two. The finiteness of the set of solutions $(m,n)$ to the equation $a^m=b^n+r$ had long been known, and Pallai himself gave some quantitative results on this using Siegel's Theorem. For finiteness alone without quantification, Bennett cites this 1918 Polya paper. My source for all of this is Bennett's paper.

Answer by Felipe Voloch for Does 2^m = 3^n + r have finitely many solutions for every r?

2010-03-20T02:41:56Z

Yes, it is true that this kind of equation ax+by=c, where a,b,c are non-zero and fixed and x,y are allowed to only have prime factors in a finite set, has only finitely many solutions. This is a special case of Siegel's theorem on integral points on curves.

Your second question may be unknown in the generality you pose. It is interesting that it holds. A remark: if there is a solution to $2^m = 3^n + r$, then $2^{m+k\phi(M)} \equiv 3^{n+k\phi(M)} + r (\mod M)$ for all $k,M$ if $(M,6)=1$, so if $M$ exists in this case, then $(M,6)>1$. If there is no solution to the equation $2^m = 3^n + r$, then the existence of $M$ (with $N=0$) is a special case of a conjecture of Skolem.

T. Skolem: Anwendung exponentieller Kongruenzen zum Beweis der Unlösbarkeit gewisser diophantischer Gleichungen., Avh. Norske Vid. Akad. Oslo, 12 (1937), 1–16.

Another comment. There are no solutions when $r=11$ but $M=8$ doesn't work since $2^2 \equiv 3^2 + 11 \mod 8$. $M(11)=205$. (Edit: $M(11)=8$ is OK. I misunderstood the definition, see comments)