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Old 2009-04-10, 11:32   #4
R.D. Silverman
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Nov 2003

22·5·373 Posts

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.
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