20090410, 01:20  #1 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
41·229 Posts 
Simple Diophantine equation
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 20090410 at 02:12 
20090410, 01:56  #2 
Aug 2006
3^{2}·5·7·19 Posts 
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 20090410 at 02:36 
20090410, 02:38  #3 
Aug 2006
3^{2}×5×7×19 Posts 
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).

20090410, 11:32  #4  
Nov 2003
2^{2}·5·373 Posts 
Quote:
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. 

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