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For every positive integer $n$ and every $\alpha\in\mathbb{Z_n}$, we have $\alpha^n=\alpha^{n-\phi(n)}$

I have found this exercise in Victor Shoup's A Computational Introduction to Number Theory and Algebra ; it's exercise 2.31 on page 35. Sadly, I cannot prove it, so I would appreciate some help.

What I tried was to somehow mimick the proof of Euler's Theorem which uses the fact that multiplying all elements of $\mathbb{Z_n}^*$ by one of the elements of the same set maps precisely the entire set. I also noticed that $ n-\phi(n)$ is precisely the number of non-invertible elements of $\mathbb{Z_n}$.

PhantomR
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  • What is $\phi$? Is it Euler's totient function? – José Carlos Santos Sep 18 '19 at 11:00
  • Mainly, you need Euler's theorem : If $a$ and $n$ are coprime, then $$a^{\varphi(n)}\equiv 1\mod n$$ holds , where $\varphi(n)$ denotes the totient-function (also called euler-phi-function) – Peter Sep 18 '19 at 11:04
  • Googling "carmichael function" will be very helpful as well. – Peter Sep 18 '19 at 11:06
  • Yes, it is Euler's totient. I'd appreciate a hint as to how I would have to apply Euler's theorem. – PhantomR Sep 18 '19 at 11:06
  • I read a little about Carmichael's function too, but couldn't extract something useful out of it, not to mention the book does not describe the function up to this exercise (only a particular case for semi-primes). – PhantomR Sep 18 '19 at 11:07
  • See here. – Bill Dubuque Sep 18 '19 at 13:15
  • @Bill Dubuque Sadly, I am unable to apply the result to this problem. Maybe you wanted me to somehow use the proof idea there, but I could not since by writing $n=q\phi+r$ and noticing that $q\geq 1$, I reach the congruence $a^{(q-1)\phi+r}(a^\phi-1)\equiv 0 \pmod{n}$, to which the lemma in the chosen answer does not seem to be easily appliable. I'd really appreciate some more hints, I've been struggling with this for days. – PhantomR Sep 19 '19 at 16:34
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    Apply the theorem to $, a^{\large n-\phi}(a^{\large \phi}-1).,$ We have $, e = n-\phi,$ so we need to show $,p^k\mid\mid n,\Rightarrow, e\ge k.\ $ Let $, n = p^k n',\ p\nmid n'.,$ Then $, n-\phi = p^k n' - (p!-!1)p^{\large k-1}\phi(n') = p^{\large k-1}(pn'-(p!-!1)\phi(n')).,$ Can you prove that is $\ge k?\ \ $ – Bill Dubuque Sep 19 '19 at 17:25
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    Wow, that's amazing! I can finish it from there since $p>p-1>0$ and $n'\geq \phi(n')$ implies that the paranthesis is $\geq 1$ (as it is an integer $>0$), so $e \geq p^{k-1} \geq 2^{k-1}$ and I can prove (e.g by induction) that $2^{k-1}\geq k,\forall k\geq 1$. Thank you very much! – PhantomR Sep 19 '19 at 20:13
  • @PhantomR You can post an answer and accept it. That may help future readers. – Bill Dubuque Sep 23 '19 at 00:48
  • @BillDubuque I'm sorry for the delay, I actually read your reply, but forgot to do something about it. I'll write a full proof soon, in a couple of hours :). Thanks a lot once again! – PhantomR Sep 28 '19 at 19:17

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Let $\phi(n) := \phi$ and also $a := \alpha$ (in order to simplify the typing of the solution).

We can rearrange the equality like this, following Bill Dubuque's advice:

$$a^{n-\phi}(a^\phi -1)=0$$

Now we want to apply Bill Dubuque's theorem from here with $m:= n$, $e:=n-\phi$, $f:=\phi$, which will solve the problem.

Noting that the condition $\phi(p_{i}^{e_i})\mid f=\phi$ is verified for all $i$ because of the multiplicity of the $\phi(\cdot)$ function, we only have to prove that $e\geq e_i,\forall i$.

For this, let $k$ be one of the $e_i$ corresponding to the prime $p:=p_i$, i.e $n=p^kn'$ and $p \nmid n'$.

We now have, following another one of Bill Dubuque's advices :), $$e = n-\phi = p^kn'- \phi(p^kn') \stackrel{\gcd(p,n')=1}{=} p^kn'-\phi(p^k)\phi(n')$$ $$= p^kn' - (p-1)p^{k-1}\phi(n')=p^{k-1}(pn'-(p-1)\phi(n'))$$

Noting that $p>(p-1)>0$ and $n'\geq \phi(n')> 0$ (the latter can be proved by using the expansion of $\phi(n)$ derived from the prime decomposition of $n$), this implies that $pn'>(p-1)\phi(n')$, so the previous paranthesis is $>0$. Since the paranthesis also has an integer value, it must be $\geq 1$, which implies $e\geq p^{k-1}$.

By induction, for $p\geq 2$ and $k-1\geq 0$, we have $p^{k-1}\geq k$, so we have proved that $e\geq k$, which completes our proof.

PhantomR
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