A positive integer k is self invertible modulo p iff. $\displaystyle k=\pm1 \ \mbox{(mod p)} $

How to prove this?

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- Jul 16th 2010, 01:39 PMdwsmithSelf-Invertible
A positive integer k is self invertible modulo p iff. $\displaystyle k=\pm1 \ \mbox{(mod p)} $

How to prove this? - Jul 16th 2010, 02:37 PMmelese
I assume you meant $\displaystyle k\equiv\pm1(mod\ p)$.

By definition an integer $\displaystyle b $ is called an inverse of $\displaystyle k $ modulo a prime $\displaystyle p $ if $\displaystyle b $ satisfies $\displaystyle kx\equiv1(mod\ p)$.

Here $\displaystyle k $ is an inverse of itself, therefore $\displaystyle k^2\equiv1(mod\ p)$ ... - Jul 16th 2010, 02:38 PMtonio

I pressume p is a prime, and then $\displaystyle k$ is self-invertible modulo p iff

$\displaystyle k^2=1\!\!\pmod p\iff (k-1)(k+1)=0\!\!\pmod p\iff k-1=0\!\!\pmod p\,\,or\,\,\,k+1=0\!\1\pmod p$

The second biconditional following from the fact that p is prime.

Tonio