# A proof in abstract algebra

• Mar 2nd 2010, 10:22 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 2nd 2010, 11:46 PM
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 $\displaystyle x, y$ such that $\displaystyle cx \equiv cy \equiv b \text{ mod } p$ then this means that $\displaystyle p|(cx-cy)$, by definition of congruence. So, $\displaystyle p|c(x-y)$. As $\displaystyle c \not\equiv 0 \text{ mod } p$ then $\displaystyle p|(x-y)$. Thus, $\displaystyle x \equiv y \text{ mod } p$.
• Mar 8th 2010, 03:59 PM
rainyice
Quote:

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

you are very helpful ~ thank you ^_^