Simple Diophantine equation

Batalov · 2009-04-10, 01:20

$a^3 = b^5 + 100$
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.

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:

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).

R.D. Silverman · 2009-04-10, 11:32

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.

2009-04-10, 01:20	#1
Batalov "Serge" Mar 2008 Phi(4,2^7658614+1)/2 2×3×5×311 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 1756₁₆ 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

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