The question asks you to find if a statement is true, if it is, prove it. If it is not, give a counterexample.

If a^{2}=b^{2}(mod p), then a=b (mod p), where p is prime.

Note the underlined equal signs are actually meant to be congruence signs. Is there a way to insert congruence signs in this forum?

I based my proof on the properties of congruences which states that if a=b (mod m), and c=d (mod m), then ac=bd (mod m).

a=b (mod m)

a = b + mk

(a)(a) = (b + mk)(b + mk)

a^{2}= b^{2}+ 2mbk + m^{2}k^{2}

a^{2}= b^{2}+ m(2bk + mk^{2}) 2bk + mk^{2}is an integer, say t

a^{2}= b^{2}+ mt

Therefore a^{2}=b^{2}(mod m)

I personally don't see why this proof would not hold if the modulus is prime. Then again, I'm not great with proofs. Can I stick a "p" in place of "m" in my proof and call it a day? Or does the fact the modulus is prime complicate things?