Self-Invertible

**dwsmith** · July 16th 2010, 01:39 PM

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

How to prove this?

**melese** · July 16th 2010, 02:37 PM

Originally Posted by dwsmith

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

How to prove this?

I assume you meant $k\equiv\pm1(mod\ p)$ .

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

Here $k$ is an inverse of itself, therefore $k^2\equiv1(mod\ p)$ ...

**tonio** · July 16th 2010, 02:38 PM

Originally Posted by dwsmith

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

How to prove this?

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

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

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

Tonio

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