Does 2^m = 3^n + r have finitely many solutions for every r? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T03:51:33Z http://mathoverflow.net/feeds/question/18817 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/18817/does-2m-3n-r-have-finitely-many-solutions-for-every-r Does 2^m = 3^n + r have finitely many solutions for every r? Dave R 2010-03-20T02:00:19Z 2010-03-20T12:21:24Z <p>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.</p> <p>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?</p> <p>$$\forall r,\ \exists M,\ \exists N,\ \forall m,n \ge N,\ \ 2^m \not\equiv 3^n + r \pmod{M}$$</p> <p>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.</p> <p>$M(1) = 8$, $M(3) = 3$, $M(5) = 1088$, $M(7) = 1632$, $M(9) = 3$, $M(11) = 8$.</p> http://mathoverflow.net/questions/18817/does-2m-3n-r-have-finitely-many-solutions-for-every-r/18818#18818 Answer by Jonas Meyer for Does 2^m = 3^n + r have finitely many solutions for every r? Jonas Meyer 2010-03-20T02:41:12Z 2010-03-20T06:58:20Z <p>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 [<strong>Edit:</strong> see below]. Here's a paper by Bennett from 2001 that shows more: <a href="http://www.math.ubc.ca/~bennett/B-CJM-Pillai.pdf" rel="nofollow">http://www.math.ubc.ca/~bennett/B-CJM-Pillai.pdf</a>. 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.</p> <p><strong>Edit:</strong> What Stroeker and Tijdeman actually did was sharpen the result by showing that except when $r$ is in <code>$\{-1,5,13\}$</code>, 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 <a href="http://www.springerlink.com/content/x5n7q1p0875726u6/" rel="nofollow">1918 Polya paper</a>. My source for all of this is Bennett's paper. </p> http://mathoverflow.net/questions/18817/does-2m-3n-r-have-finitely-many-solutions-for-every-r/18819#18819 Answer by Felipe Voloch for Does 2^m = 3^n + r have finitely many solutions for every r? Felipe Voloch 2010-03-20T02:41:56Z 2010-03-20T12:21:24Z <p>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. </p> <p>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.</p> <p>T. Skolem: Anwendung exponentieller Kongruenzen zum Beweis der Unlösbarkeit gewisser diophantischer Gleichungen., Avh. Norske Vid. Akad. Oslo, 12 (1937), 1–16.</p> <p>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)</p>