# Self-Invertible

• Jul 16th 2010, 01:39 PM
dwsmith
Self-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 PM
melese
Quote:

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

How to prove this?

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 PM
tonio
Quote:

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

How to prove this?

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