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 2009-04-10, 01:20 #1 Batalov     "Serge" Mar 2008 Phi(4,2^7658614+1)/2 2×3×1,571 Posts 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
 2009-04-10, 01:56 #2 CRGreathouse     Aug 2006 32×5×7×19 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 2009-04-10 at 02:36
2009-04-10, 02:38   #3
CRGreathouse

Aug 2006

32·5·7·19 Posts

Quote:
 Originally Posted by Batalov 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).

2009-04-10, 11:32   #4
R.D. Silverman

Nov 2003

22·5·373 Posts

Quote:
 Originally Posted by Batalov $a^3 = b^5 + 100$ 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.

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