# A proof in abstract algebra

• Mar 2nd 2010, 11:22 PM
rainyice
A proof in abstract algebra
Prove that if p is a prime and c is not congruent 0 (mod p), then cx is congruent b (mod p) has a unique solutin modulo p. That is, a solution exists, and any two solutions are congruent modulo p.
• Mar 3rd 2010, 12:46 AM
Swlabr
Quote:

Originally Posted by rainyice
Prove that if p is a prime and c is not congruent 0 (mod p), then cx is congruent b (mod p) has a unique solutin modulo p. That is, a solution exists, and any two solutions are congruent modulo p.

If there exists $x, y$ such that $cx \equiv cy \equiv b \text{ mod } p$ then this means that $p|(cx-cy)$, by definition of congruence. So, $p|c(x-y)$. As $c \not\equiv 0 \text{ mod } p$ then $p|(x-y)$. Thus, $x \equiv y \text{ mod } p$.
• Mar 8th 2010, 04:59 PM
rainyice
Quote:

Originally Posted by Swlabr
If there exists $x, y$ such that $cx \equiv cy \equiv b \text{ mod } p$ then this means that $p|(cx-cy)$, by definition of congruence. So, $p|c(x-y)$. As $c \not\equiv 0 \text{ mod } p$ then $p|(x-y)$. Thus, $x \equiv y \text{ mod } p$.

you are very helpful ~ thank you ^_^