Prove that if a ^ 2 ≡ b ^ 2 (mod p), then a ≡ ± b (mod p).

Question

hello could help me with this problem I used a theorem but I do not know if esat well

Answer 1

Use that $$a^2-b^2=(a-b)(a+b)$$

Answer 2

Here is a proof for $p$ prime: If $a^2\equiv b^2\mod p$, then equivalently $(a-b)(a+b)\equiv 0\mod p$. But the residue class ring ${\Bbb Z}_p$ is a field and so has no zero divisors. This gives $a\equiv\pm b\mod p$.

For $p$ not prime, the assertion fails: $2^2 \equiv 4^2\mod 12$, but not $4\equiv \pm 2\mod 12$.