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Thread: congruence

  1. #1
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    congruence

    a) If a^p = b^p mod p, prove that a = b mod p

    b) If a a = b mod p, prove that a^p = b^p mod p
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  2. #2
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    #1: $\displaystyle b \equiv b^{p} \equiv a^{p} \equiv aa^{p - 1} \equiv \hdots \ (\text{mod } p)$ (Hint: Fermat's theorem)

    #2: You can use the fact that if $\displaystyle a \equiv b \ (\text{mod }p)$ and $\displaystyle c \equiv d \ (\text{mod }p)$ then $\displaystyle ac \equiv bd \ (\text{mod } p)$. Then it's just a matter of considering $\displaystyle c = a$ and $\displaystyle d = b$
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    Quote Originally Posted by o_O View Post
    #1: $\displaystyle b \equiv b^{p} \equiv a^{p} \equiv aa^{p - 1} \equiv \hdots \ (\text{mod } p)$ (Hint: Fermat's theorem)

    #2: You can use the fact that if $\displaystyle a \equiv b \ (\text{mod }p)$ and $\displaystyle c \equiv d \ (\text{mod }p)$ then $\displaystyle ac \equiv bd \ (\text{mod } p)$. Then it's just a matter of considering $\displaystyle c = a$ and $\displaystyle d = b$

    trying to use fermats theorem but still cant prove it
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    Quote Originally Posted by bigb View Post
    a) If a^p = b^p mod p, prove that a = b mod p
    Do what o_O did.
    Fermat's little theorem says $\displaystyle a^p \equiv a$ and $\displaystyle b^p \equiv b$.
    Therefore we can replace $\displaystyle a^p$ by $\displaystyle a$ and $\displaystyle b^p$ by $\displaystyle b$ since they are congruent.
    This leaves us with $\displaystyle a\equiv b$.
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    Quote Originally Posted by ThePerfectHacker View Post
    Do what o_O did.
    Fermat's little theorem says $\displaystyle a^p \equiv a$ and $\displaystyle b^p \equiv b$.
    Therefore we can replace $\displaystyle a^p$ by $\displaystyle a$ and $\displaystyle b^p$ by $\displaystyle b$ since they are congruent.
    This leaves us with $\displaystyle a\equiv b$.
    Can anyone work this out.. really cant seem to solve this
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    Fermat's theorem: $\displaystyle {\color{red}a^{p-1} \equiv 1 \ (\text{mod } p)}$

    $\displaystyle b \equiv b^{p} \equiv a^{p} \equiv a{\color{red}a^{p - 1}} \equiv a{\color{red}(1)} \ (\text{mod } p)$
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  7. #7
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    Quote Originally Posted by bigb View Post
    a) If a^p = b^p mod p, prove that a = b mod p

    b) If a a = b mod p, prove that a^p = b^p mod p
    I made a mistake in typing out the problem for part b if that makes a differnece...it should be a^p = b^p mod p, prove that a^P = b^p mod p^2
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