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Thread: A proof in abstract algebra

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    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.
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  2. #2
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by rainyice View Post
    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, $\displaystyle x \equiv y \text{ mod } p$.
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    Quote Originally Posted by Swlabr View Post
    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 ^_^
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