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Math Help - 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: 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 a \equiv b \ (\text{mod }p) and c \equiv d \ (\text{mod }p) then ac \equiv bd \ (\text{mod } p). Then it's just a matter of considering c = a and d = b
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    Quote Originally Posted by o_O View Post
    #1: 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 a \equiv b \ (\text{mod }p) and c \equiv d \ (\text{mod }p) then ac \equiv bd \ (\text{mod } p). Then it's just a matter of considering c = a and d = b

    trying to use fermats theorem but still cant prove it
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  4. #4
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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 a^p \equiv a and b^p \equiv b.
    Therefore we can replace a^p by a and b^p by b since they are congruent.
    This leaves us with a\equiv b.
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  5. #5
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    Quote Originally Posted by ThePerfectHacker View Post
    Do what o_O did.
    Fermat's little theorem says a^p \equiv a and b^p \equiv b.
    Therefore we can replace a^p by a and b^p by b since they are congruent.
    This leaves us with a\equiv b.
    Can anyone work this out.. really cant seem to solve this
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  6. #6
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    Fermat's theorem: {\color{red}a^{p-1} \equiv 1 \ (\text{mod } p)}

    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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