Prove or find a counterexample: Let p be a prime, and a and b positive integers. If a^2 ≡ b^2 (mod p) then a ≡ b (mod p).
I know that: If a and b are integers, we say that they are congruent modulo d iff b − a is a multiple of d. Equivalently, a and b are congruent mod d iff a (mod d) = b (mod d). Rewrite this as: a ≡ b (mod d).
I also know if p is prime, then GCD(a, p) = 1, GCD(b, p) = 1, but I'm not entirely sure if any of this information is useful or not
Think about it this way, if $a^2 \equiv b^2 (p)$ then we must also have that $a^2-b^2 \equiv 0 (p)$.
If you can answer the following questions you should obtain your answer:
Can we factor $a^2-b^2$?
What does $p$ being prime tell us about it ability to divide the factors of $a^2-b^2$?