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Batalov 2009-04-10 01:20

Simple Diophantine equation
 
[tex]a^3 = b^5 + 100[/tex]
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 ...[I]too[/I] simple.

CRGreathouse 2009-04-10 01:56

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.

CRGreathouse 2009-04-10 02:38

[QUOTE=Batalov;168698]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 ...[I]too[/I] simple.[/QUOTE]

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).

R.D. Silverman 2009-04-10 11:32

[QUOTE=Batalov;168698][tex]a^3 = b^5 + 100[/tex]
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 ...[I]too[/I] simple.[/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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