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 $\displaystyle k^2\equiv1(mod\ p)$ ...