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  1. MHB Master
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    #1
    Hey!!!

    I have the following exercise:

    If $p$ is prime, $p \nmid a$, $p \nmid b$, prove that $$a^p \equiv b^p \pmod p \Rightarrow a^p \equiv b^p \pmod {p^2}$$

    My idea is the following:
    $$ a^p \equiv b^p \pmod p $$
    $$ a^p \equiv a \pmod p $$
    $$ b^p \equiv b \pmod p $$

    $$ a \equiv b \pmod p \Rightarrow a=b+kp, k \in \mathbb{Z}$$

    $$ a^p=(b+kp)^p=\sum_{m=0}^{p} \binom{p}{m} (kp)^mb^{p-m}=\binom{p}{0}b^p+\binom{p}{1}(kp)b^{p-1}+ \dots + \binom{p}{p}(kp)^p= \\ b^p+kp^2b^{p-1}+ \frac{1}{2}(p-1)p^3b^{p-2}k^2+\dots + k^pp^p$$

    $$ p^2 \mid kp^2b^{p-1}+ \frac{1}{2}(p-1)p^3b^{p-2}k^2+\dots + k^pp^p
    \Rightarrow p^2 \mid a^p-b^p $$

    Is this correct??

  2. MHB Master
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    #2
    That looks correct.

  3. MHB Master
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    #3 Thread Author
    Quote Originally Posted by Deveno View Post
    That looks correct.
    Do we not have to prove that $$kb^{p-1}+ \frac{1}{2}(p-1)pb^{p-2}k^2+\dots + k^pp^{p-2} \in \mathbb{Z} $$??

  4. MHB Seeker
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    #4
    Quote Originally Posted by mathmari View Post
    Do we not have to prove that $$kb^{p-1}+ \frac{1}{2}(p-1)pb^{p-2}k^2+\dots + k^pp^{p-2} \in \mathbb{Z} $$??
    Yep. We do.
    Can you?

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    #5
    Quote Originally Posted by I like Serena View Post
    Yep. We do.
    Can you?
    Consider that each of the binomial coefficients is of the form:
    $$\binom p k = \frac{p \cdot (p-1) \cdot ... \cdot (p-k+1)}{1\cdot 2 \cdot ... \cdot k}$$
    It is given that this is an integer.
    So each of those factors in the denominator come back in the numerator somehow.
    Can any of them contain a factor that divides $p$?

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    #6
    One can prove that:

    $\displaystyle \binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}$

    which furnishes an easy inductive proof that for all $0 \leq k \leq n$ that $\displaystyle \binom{n}{k}$ is an integer, given:

    $\displaystyle \binom{n}{0} = \binom{n}{n} = 1$, for all $n \in \Bbb Z^+$.

    Then the only fact we need in the expansion of $(b + kp)^p$ is that $\displaystyle p|\binom{p}{1}$

    since $p^2$ occurs in all the other terms of the expansion, no matter what the coefficient is.

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