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

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 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.

(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 [itex] \displaystyle p_{i}^{e_{i}-1} \geq e_{i}).
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