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Math Help - [SOLVED] Prove that a^p is congruent to b^p (mod p^2) if the same is true mod p

  1. #1
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    [SOLVED] Prove that a^p is congruent to b^p (mod p^2) if the same is true mod p

    For any prime p, if a^p\equiv b^p\;(\text{mod } p), prove that a^p\equiv b^p\;(\text{mod } p^2).
    I have found this quite challenging!

    I thought maybe we could divide it into two cases, where p\big|a,b and p\not\big|a,b.

    If we consider Euler's function \phi(n), then we have, for p\not\big|a,b,

    b^{\phi(p^2)}\equiv a^{\phi(p^2)}\equiv 1\;(\text{mod }p^2).

    Since \phi(p^2)=p^2-p, then

    b^{p^2-p}\equiv a^{p^2-p}\equiv 1\;(\text{mod }p^2).

    So we could say that

    a^{p^2}\equiv a^p\;(\text{mod }p^2) and b^{p^2}\equiv b^p\;(\text{mod }p^2).

    And we can say a bunch of other things, too, none of which seem to get me closer to a solution.

    Any help would be much appreciated!
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  2. #2
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    Quote Originally Posted by hatsoff View Post
    I have found this quite challenging!

    I thought maybe we could divide it into two cases, where p\big|a,b and p\not\big|a,b.

    If we consider Euler's function \phi(n), then we have, for p\not\big|a,b,

    b^{\phi(p^2)}\equiv a^{\phi(p^2)}\equiv 1\;(\text{mod }p^2).

    Since \phi(p^2)=p^2-p, then

    b^{p^2-p}\equiv a^{p^2-p}\equiv 1\;(\text{mod }p^2).

    So we could say that

    a^{p^2}\equiv a^p\;(\text{mod }p^2) and b^{p^2}\equiv b^p\;(\text{mod }p^2).

    And we can say a bunch of other things, too, none of which seem to get me closer to a solution.

    Any help would be much appreciated!

    At first sight this looks quite trivial, but let's see:

    a^p = b^p (mod p) ==> a = b (mod p), since a^p = a (mod p) for any

    integer a ==> a = b + kp ==> a^p = (b + kp)^p = b^p + (sum of

    numbers all of which are divisible by p^2) ==> a^p = b^p (mod p^2)

    The only non quite trivial thing here, imo, is why in that sum all the

    numbers are divisible by p? Well, the first summand there is p(kp) and the

    following ones are (binomial coefficient)*(kp)^i, with i >= 2, so...

    Tonio

    Pd. Indeed not "quite trivial"
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  3. #3
    MHF Contributor Bruno J.'s Avatar
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    If a^p\equiv b^p then a \equiv b (because n \equiv n^p \mod p).

    Hence b = kp + a for some integer k. Then

    b^p -a^p= (kp+a)^p -a^p= (kp)^p +{p \choose 1}(kp)^{p-1}a + ... + {p \choose 1}kpa^{p-1} .

    Note that every term in this sum is divisible by p^2 because {p \choose j} \equiv 0 \mod p for 0<j<p. Hence p^2 \mid (b^p-a^p) \Rightarrow a^p \equiv b^p \mod p^2. \square
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