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I'm having a tough time with proving this statement. I've been able to do this so far: We have that $p^n \mid a - b$ and also that $p \mid a^p - a$ and $p \mid b^p - b$. Therefore $p^nk = a - b$ for some $k$, $pj = a^p - a$ for some $j$, and $pi = b^p - b$ for some $i$. Therefore $a = a^p - pj$ and $b = b^p - pi$ so $p^nk = a^p - pj - b^p + pi$, and therefore $p(p^{n - 1}k + j - i) = a^p - b^p$, and therefore $a^p \equiv b^p \mod p$. But this is just a basic property of modular arithmetic and nothing new. I can't figure out how I'm supposed to 'scale' the modulo up by $p^n$. Could I have some help to get me on the right track? Thanks for your time.

deb
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1 Answers1

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First note that: $$ p^n|a-b\Rightarrow p|a-b $$ So $a\equiv b (\mod p)$ wich also gives us $a^k\equiv b^k(\mod p)$ for all k. Now consider: $$ a^n-b^n=(a-b)(a^{p-1}+ba^{p-2}+\cdots+b^{p-2}a+b^{p-1})=(a-b)\Big(\sum_{k=0}^{p-1}a^kb^{p-1-k} \Big)\quad (1) $$ Notice that for each k, $a^kb^{p-1-k}\equiv b^k b^{p-1pk}(\mod p)\equiv b^{k+p-1-k}(\mod p)\equiv b^{p-1}(\mod p)\equiv 1 (\mod p)$ by FLT. So the sum part becomes: $$ \sum_{k=0}^{p-1}a^kb^{p-1-k} \equiv \sum_{k=0}^{p-1} 1 (\mod p)=0(\mod p) $$ So by definition $\displaystyle p|\sum_{k=0}^{p-1}a^kb^{p-1-k} $. So by the above:

  1. $\displaystyle \sum_{k=0}^{p-1}a^kb^{p-1-k}=k_1 p$
  2. $a-b=k_2 p^n $ by hypothesis

So (1) becomes: $$ a^n-b^n=k_2p^nk_1p=k_1k_2p^{n+1} \Rightarrow a^n\equiv b^n(\mod p^{n+1}) $$

Bigalos
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