Sunday, 6 March 2011

nt.number theory - Does 2^m = 3^n + r have finitely many solutions for every r?

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 2m=3n+r, then 2m+kphi(M)equiv3n+kphi(M)+r(modM) 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 2m=3n+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 22equiv32+11mod8. M(11)=205. (Edit: M(11)=8 is OK. I misunderstood the definition, see comments)

No comments:

Post a Comment