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?
Hmm, when I originally attempted to prove this I went down this road, but stalled. If you brought this up, it must be significant. This was my approach...
If
then
so either or
if
then
which means
if
then
which means , but since -b is not neccesarily congruent to b (mod p) we cannot conclude that .
So that's where I stalled...
If you'd like, you can post the correct answer once you know it so that other people who will read the thread later may have the complete picture.
Also, the converse of an implication should not be confused with the negation. The converse of A -> B is B -> A. The negation applies to any proposition, and the negation ~(A -> B) of an implication is equivalent to A /\ ~B. Proving the negation of a statement indeed disproves that statement.