1. Modules

Let p be any odd prime number and set A = {0, 1, 2, . . . , p − 1}. Show there are exactly
two solutions x belongs to 2 A which satisfy (x^2) ≡ 1 (mod p).

2. Arithmetic mod p

Hello smithhall
Originally Posted by smithhall
Let p be any odd prime number and set A = {0, 1, 2, . . . , p − 1}. Show there are exactly
two solutions x belongs to 2 A which satisfy (x^2) ≡ 1 (mod p).
I assume that the 2 shouldn't be there!

$\displaystyle x^2 \equiv 1 \mod p$

$\displaystyle \Rightarrow x^2 = np+1,\, n \in \mathbb{N}$

$\displaystyle \Rightarrow (x+1)(x-1) = np$

$\displaystyle \Rightarrow x = 1, n = 0, \text{ or }x+1 = \frac{np}{x-1}$

$\displaystyle \Rightarrow (x-1)|n$, since $\displaystyle x-1<p$ and $\displaystyle p$ is prime

$\displaystyle \Rightarrow n = m(x-1), m\in\mathbb{N}$

$\displaystyle \Rightarrow x+1 = mp \equiv 0 \mod p$

$\displaystyle \Rightarrow x \equiv -1 \mod p$

$\displaystyle \Rightarrow x = p-1$, since $\displaystyle 0<x\le (p-1)$

So if $\displaystyle p>2$, there are just two distinct solutions: $\displaystyle x = 1, p-1$