The solution of x2 ≡ 1 mod p is x ≡ ±1 mod p

If \(x^2 ≡ 1 \mod p\), then:

$$ p|(x^2-1) $$$$ p|(x-1)(x+1) $$

Using [this lemma], p|(x-1) or p|(x+1):

$$ x ≡ ±1 \mod p $$

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