Problem 40: PEN, Diophantine Equations, Problem 14

by henderson, Jul 17, 2016, 3:41 PM

$\color{red}\bf{Problem \ 40}$ Show that the equation
$x^2+y^5=z^3$ has infinitely many solutions in integers $x,y,z$ for which $xyz\not= 0.$
(Canada 1991) $($ My solution

$\color{red}\bf{My \ solution}$ Firstly, observe that
$10^2+3^5=7^3.$ Then, replacing $x=10a^{15}, y=3a^6$ and $z=7a^{10}$ $($ where $a$ is an arbitrary integer different from $0$ $),$ we are done.

$)$

This post has been edited 8 times. Last edited by henderson, Sep 28, 2016, 12:36 PM

Diophantine Equations

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Problem 40: PEN, Diophantine Equations, Problem 14

by henderson, Jul 17, 2016, 3:41 PM

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