Special root

Number Theory Level 3

3 2 − 1 3 = 2 \large \sqrt[3]{3^2 - 1} = 2 Are there any other positive integers a ≥ 2 , b ≥ 2 , c ≥ 2 a \geq 2, b \geq 2, c \geq 2 such that c b − 1 a \large \sqrt[a]{c^b - 1} is a positive integer?

No Yes

Arjen Vreugdenhil by Arjen Vreugdenhil , 44, USA

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Arjen Vreugdenhil
Mar 23, 2020

No, there are no other such positive integers. This follows immediately from Catalan's conjecture, that there is only one pair of perfect powers that differ by one (namely 8 and 9).

This is longer merely a conjecture but was proven in 2002 by Preda Mihăilescu.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

And what about a=3,b=1,c=2?

Dan Czinege - 1 year, 2 months ago

Log in to reply

The question specifically asks for a ≥ 2 , b ≥ 2 , c ≥ 2 a \ge 2, b \ge 2, c \ge 2 .

Elijah L - 1 year, 2 months ago

Log in to reply

Well, he must have added this assumption, because when I was solving this problem it wasn´t written there. :) But now its fine.

Dan Czinege - 1 year, 2 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...