Science:Math Exam Resources/Courses/MATH312/December 2008/Question 06 (d)

MATH312 December 2008

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Question 06 (d)

The purpose of this problem is to prove the following tThe 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 prime-power 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 prime-power factorization of m. (d) Complete the proof of Theorem 1 above

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
Do parts (a) to (c) for each prime factor.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Notice that parts (a) to (c) held for an arbitrary prime factor of m. We repeat the above for each such prime factor. Since they are all distinct primes, the fact that $\displaystyle p_{i}^{e_{i}}\mid (a^{m}-a^{m-\phi (m)})$ holds for each $\displaystyle i\in \{1,..,k\}$ will together imply that via the Chinese Remainder Theorem $\displaystyle m=\prod _{i=1}^{k}p_{i}^{e_{i}}\mid (a^{m}-a^{m-\phi (m)})$ and this completes the proof.

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