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Old 2014-07-14, 23:35   #3
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"Forget I exist"
Jul 2009

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Originally Posted by Zeta-Flux View Post
Better than that, the Wieferich condition is equivalent to 2^{(p-1)/2} \equiv \pm 1\pmod{p^2}. This makes it slightly easier to test for the condition.

The only place I know of where the Wieferich condition is really useful is in the first case of Fermat's last theorem and in the solution to Catalan's conjecture.
Okay, Thanks for that, I just thought it might be useful for trying to pin down what n are possible to create such p instead of trying any p ( which as far as I know, and yes I don't know much if any, is how it works).

Last fiddled with by science_man_88 on 2014-07-14 at 23:39
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