Question

Prove that if p is prime and 0 ≤ x < p and x2 ≡ 1 (mod p), then either x = 1 or x = p-1. Hint: Begin by observing that x2 ≡ 1 (mod p) if and only if x2 - 1 ≡ 0 (mod p). Then factor.

Answer 1

To prove that if p is prime and 0 ≤ x < p and x² ≡ 1 (mod p), then either x = 1 or x = p-1, we can start by observing that x² ≡ 1 (mod p) if and only if x² - 1 ≡ 0 (mod p). From here, we can factor the equation as (x - 1)(x + 1) ≡ 0 (mod p). Therefore, either x = 1 or x = p-1.

Step-by-step explanation:

To prove that if p is prime and 0 ≤ x < p and x² ≡ 1 (mod p), then either x = 1 or x = p-1, we can start by observing that x² ≡ 1 (mod p) if and only if x² - 1 ≡ 0 (mod p). From here, we can factor the equation as (x - 1)(x + 1) ≡ 0 (mod p). Since p is prime, it means that p cannot divide both (x - 1) and (x + 1) simultaneously, except when x ≡ ±1 (mod p). Therefore, either x = 1 or x = p-1.