- **Puzzles**
(*https://www.mersenneforum.org/forumdisplay.php?f=18*)

- - **Simple Diophantine equation**
(*https://www.mersenneforum.org/showthread.php?t=11708*)

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. |

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. |

[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). |

[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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