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I'm working through the book An Introduction to the Theory of Numbers by Ivan M. Niven and I'm having trouble with this exercise:

Chapter 2.3, Problem 44. Prove the following generalization of Euler's theorem: $$ a^m \equiv a^{m-\phi(m)} \pmod{m} $$ for any integer $a$.

calculatormathematical
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thlayli
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  • Are you familiar with the Chinese remainder theorem? – Tobias Kildetoft Feb 18 '16 at 07:23
  • Yes; this question is in the section on CRT, so I figured I might have to use that. -- but would welcome some hint on how to start. – thlayli Feb 18 '16 at 07:24
  • Recall that the CRT says that given some set of congruences mod pairwise coprime numbers, there is a unique solution mod the product of these. Try to apply this to the factorization of $m$ into prime powers. – Tobias Kildetoft Feb 18 '16 at 07:25
  • So something like $x\equiv a^{m-\phi(m)}$ (mod $p_i^{e_i}$), where $m = \prod p_i^{e_i}$? – thlayli Feb 18 '16 at 07:35
  • Right. Can you show that $a^m$ is a solution to that? – Tobias Kildetoft Feb 18 '16 at 07:36
  • I'm trying to use the construction of the solution the book used in the proof of the CRT. But how would I find the inverse of $\frac{m}{p_i^{e_i}}$ mod ${p_i^{e_i}}$? – thlayli Feb 18 '16 at 07:41
  • You don't need this. You just need to use that if some $x$ satisfies all those congruences, then any other solution will be congruenct to $x$ mod $m$. – Tobias Kildetoft Feb 18 '16 at 07:42
  • Maybe I'm missing something obvious, but I can't figure out how to show $a^m$ is a solution to that. It doesn't seem obvious that $p_i^{e_i} | a^m(1 - a^{-\phi(m)})$. – thlayli Feb 18 '16 at 07:48
  • Split it into two cases depending on whether $p_i$ divides $a$ or not. Can you show it in one of those? – Tobias Kildetoft Feb 18 '16 at 07:53
  • Ah, I think I see. And in the other case, they are relatively prime, which I've already figured out. Thank you very much for your help! – thlayli Feb 18 '16 at 08:00

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Using the fundamental theorem of number theory we can write $m$ as a product of primes $$m=\prod_i p_i^{\alpha_i}.$$ By theorem 2.3 part (3) in the book, which is some sort of chinese remainder theorem, it suffices to show that $a^m\equiv a^{m-\phi(m)}$ modulo $p_i^{\alpha_i}$ for each $i$. So fix an arbitrary $i$, and write $m=p_i^{\alpha_i} m^{\prime}$, with $\operatorname{gcd}\left(m^{\prime}, p_i\right)=1$. Then by theorem 2.19, $$\phi(m)=\left(p^{\alpha_i}-p^{\alpha_i-1}\right) \phi\left(m^{\prime}\right).$$ Now consider two cases. If $\left(a, p_i\right)=1$, then $$ a^{\phi\left(p_i^{\alpha_i}\right)}=a^{\left(p^{\alpha_i}-p^{\alpha_i-1}\right)} \equiv 1 \pmod{p_i^{\alpha_i}}, $$ so $$ a^{m-\phi(m)} \equiv a^{m-\phi(m)} \cdot\left(a^{\left(p^{\alpha_i}-p^{\alpha_i-1}\right)}\right)^{m^{\prime}} \equiv a^m \pmod{p_i^{\alpha_i}} $$ and so the result holds in this case. The second case is if $p_i \mid a$. In this case, we claim that both numbers are congruent to $0$ modulo $p_i^{\alpha_i}$. Write $a=p_i a^{\prime}$ for some $a^{\prime}$, and note that both $m$ and $$\phi(m)=\left(p^{\alpha_i}-p^{\alpha_i-1}\right) \phi\left(m^{\prime}\right)$$ are divisible by $p_i^{\alpha_i-1}$. Therefore, on both sides of the congruence we have factors of $p^{p_i^{\alpha_i-1}}$. Since $p_i^{\alpha_i-1} \geq \alpha_i$, this implies that both sides are congruent to $0$.

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