mersenneforum.org  

Go Back   mersenneforum.org > Fun Stuff > Puzzles

Reply
 
Thread Tools
Old 2009-04-10, 01:20   #1
Batalov
 
Batalov's Avatar
 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2

912610 Posts
Default Simple Diophantine equation

a^3 = b^5 + 100
Is there more than one solution?




________
P.S. Argh, I only started playing with modular restrictions, and you've already looked up to 10^35.
If so, then of course I am convinced, too. Sorry, then it was ...too simple.

Last fiddled with by Batalov on 2009-04-10 at 02:12
Batalov is offline   Reply With Quote
Old 2009-04-10, 01:56   #2
CRGreathouse
 
CRGreathouse's Avatar
 
Aug 2006

134438 Posts
Default

Doubtful. No small solutions (|a^3| < 10^35) other than 7^3 = 3^5 + 100, and powers are rarely close together.

S. S. Pillai conjectures that a positive integer can be expressed as the difference of powers only finitely many ways, which suggests that finite checking is meaningful.

Last fiddled with by CRGreathouse on 2009-04-10 at 02:36
CRGreathouse is offline   Reply With Quote
Old 2009-04-10, 02:38   #3
CRGreathouse
 
CRGreathouse's Avatar
 
Aug 2006

5,923 Posts
Default

Quote:
Originally Posted by Batalov View Post
P.S. Argh, I only started playing with modular restrictions, and you've already looked up to 10^35.
If so, then of course I am convinced, too. Sorry, then it was ...too simple.
I just checked -10 million to 10 million, raised to the fifth power and added 100, and checked if the numbers were cubes (using modular restrictions to avoid taking too many cube roots).
CRGreathouse is offline   Reply With Quote
Old 2009-04-10, 11:32   #4
R.D. Silverman
 
R.D. Silverman's Avatar
 
Nov 2003

22·5·373 Posts
Default

Quote:
Originally Posted by Batalov View Post
a^3 = b^5 + 100
Is there more than one solution?




________
P.S. Argh, I only started playing with modular restrictions, and you've already looked up to 10^35.
If so, then of course I am convinced, too. Sorry, then it was ...too simple.

Note that Faltings proof of the Mordell Conjecture shows that there
are only finitely many solutions. Actually, this is like hitting a thumbtack
with a sledgehammer. Siegel's Theorem suffices to show the same thing.

Unfortunately, neither is effective. Nor would an application of the
ABC conjecture be effective.
R.D. Silverman is offline   Reply With Quote
Reply

Thread Tools


Similar Threads
Thread Thread Starter Forum Replies Last Post
Basic Number Theory 13: Pythagorean triples and a few Diophantine equations Nick Number Theory Discussion Group 2 2016-12-18 14:49
What's the basic LLR equation? jasong jasong 4 2012-02-20 03:33
Diophantine Equation flouran Math 7 2009-12-12 18:48
Diophantine Question grandpascorpion Math 11 2009-09-23 03:30
Diophantine problem philmoore Puzzles 8 2008-05-19 14:17

All times are UTC. The time now is 16:23.

Wed Sep 23 16:23:22 UTC 2020 up 13 days, 13:34, 2 users, load averages: 1.89, 1.85, 1.79

Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2020, Jelsoft Enterprises Ltd.

This forum has received and complied with 0 (zero) government requests for information.

Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation.
A copy of the license is included in the FAQ.