MATH312 December 2008
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Question 06 (b)

The purpose of this problem is to prove the following theorem.
Theorem 1. For all positive integers $\displaystyle a,m$ we have $\displaystyle a^{m}\equiv a^{m\phi (m)}\mod {m}$
Let $\displaystyle a,m$ be positive integers. For $\displaystyle m=1$, the theorem holds trivially, so we assume from now on that $\displaystyle m>1$ and write its primepower factorization as $\displaystyle m=p_{1}^{e_{1}}\cdot ...\cdot p_{k}^{e_{k}}$ for different primes $\displaystyle p_{1},...,p_{k}$ and positive integer exponents $\displaystyle e_{1},...,e_{k}$ and some positive integer k. Let $\displaystyle i\in \{1,..,k\}$ and focus on the prime power $\displaystyle p_{i}^{e_{i}}$ in the primepower factorization of m.
(b) Prove that if $\displaystyle p_{i}\mid a$ then $\displaystyle p_{i}^{e_{i}}\mid a^{m\phi (m)}$. You might want to do this as follows:
 Prove that $\displaystyle p_{i}^{e_{i}1}\mid (m\phi (m))$.
 Prove that $\displaystyle m\phi (m)\geq p_{i}^{e_{i}1}$.
 Prove that $\displaystyle p_{i}^{e_{i}}\mid p^{m\phi (m)}$ (here you may use without proof that <math> \displaystyle p_{i}^{e_{i}1} \geq e_{i}).

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint

Again the question outlines the proof fairly thoroughly.

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Solution

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We prove the sub claims as outlines in the question.
Prove that $\displaystyle p_{i}^{e_{i}}\mid (m\phi (m))$.
For this, we use the property of the phi function to see
$\displaystyle \phi (p_{i}^{e_{i}})=p_{i}^{e_{i}}(11/p_{i})=p_{i}^{e_{i}1}(p_{i}1)$
and thus by the multiplicativity of the phi function, we have
$\displaystyle m\phi (m)=\prod _{j=1}^{k}p_{j}^{e_{j}}\prod _{j=1}^{k}p_{j}^{e_{j}1}(p_{j}1)=\prod _{j=1}^{k}p_{j}^{e_{j}1}(p_{j}(p_{j}1))=\prod _{j=1}^{k}p_{j}^{e_{j}1}$
and hence $\displaystyle p_{i}^{e_{i}}\mid (m\phi (m))$.
Prove that $\displaystyle m\phi (m)\geq p_{i}^{e_{i}1}$.
The above shows that $\displaystyle m\phi (m)=\prod _{j=1}^{k}p_{j}^{e_{j}1}\geq p_{i}^{e_{i}1}$ and this completes the proof.
Prove that $\displaystyle p_{i}^{e_{i}}\mid p_{i}^{m\phi (m)}$ (here you may use without proof that $\displaystyle p_{i}^{e_{i}1}\geq e_{i}$.)
We have that
$\displaystyle p_{i}^{m\phi (m)}\geq p_{i}^{p_{i}^{e_{i}1}}\geq p_{i}^{e_{i}}$.
Therefore,
$\displaystyle p_{i}^{e_{i}}\mid p_{i}^{m\phi (m)}$.
We finish off this problem by noticing that
$\displaystyle p_{i}^{e_{i}}\mid p_{i}^{m\phi (m)}\mid a^{m\phi (m)}$.

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