From: nikl@mathematik.tu-muenchen.de (Gerhard Niklasch)
Newsgroups: sci.math
Subject: Re: That "3 prime factors" problem
Date: 24 Mar 1998 17:26:44 GMT
Keywords: S-unit equations in Z
In article <6f8ml6$1f0$1@news1.rmi.net>, Kurt Foster
reduced Another Problem to that of solving the following diophantine
equations:
|> (A1) |2^C * 5^D - 3^B| = 1
|>
|> (A2) |2^A * 3^B - 5^D| = 1
|>
|> (A3) |3^B - 5^D| = 2
|> (B1) |2^A * 3^B - 5^D| = 1
|>
|> (B2) |2^C * 5^D - 3^B| = 1
|> (C1) |2^A * 3^C -5^D| = 1
|>
|> (C2) |2^A - 3^C * 5^D| = 1
where, for the purpose at hand, it would be sufficient to know that
each of them has only finitely many solutions.
Now it is well known that the `S-unit equation' X + Y = Z in integers
X,Y,Z such that all prime divisors of XYZ come from a given finite set S
of primes has only finitely many solutions in which X,Y,Z are coprime,
and the solutions can be effectively determined. While this requires
transcendental methods in general, L.J.Alex [Comm. Algebra 4 (1976),
77--100] arrived at the complete solution for S={2,3,5,7} by elementary
methods. If you want to go a little beyond these primes, B.M.M.de Weger
[Algorithms for Diophantine Equations, CWI Tract 65, 1989, Chapter 6:
Thm.6.3 and Table II] gives the full solution set for S={2,3,5,7,11,13}.
For S={2,3,5}, the complete solution set, up to playing games with signs
and permuting summands, consists of
1 + 1 = 2
2 + 1 = 3
3 + 1 = 2^2
3 + 2 = 5
2^2 + 1 = 5
5 + 1 = 2*3
5 + 3 = 2^3
5 + 2^2 = 3^2
2^3 + 1 = 3^2
3^2 + 1 = 2*5
3*5 + 1 = 2^4
2^4 + 3^2 = 5^2
2^3*3 + 1 = 5^2
5^2 + 2 = 3^3
3^3 + 5 = 2^5
2^4*5 + 1 = 3^4
5^3 + 3 = 2^7
The application to the previous poster's exposition is left as an
exercise. :^)
(Gotta love this -- didn't take me long to hack a little GP/PARI loop
to spit these out, starting from the bounds in de Weger's theorem
since I don't have the other paper within arm's reach, _and_ to spot
Yet Another Trivial Little Buglet in pari-gp-2.0.7.alpha whilst doing
so...)
Enjoy, Gerhard
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