# Thread: Modules

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!

$x^2 \equiv 1 \mod p$

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

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

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

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

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

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

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

$\Rightarrow x = p-1$, since $0

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

Grandad