Fermat's Little Theorem, in Fermat-Euler form, states that: if $\gcd(a,m)=1$, then $a^{\varphi(m)} \equiv 1 \pmod{m}$.
Now, I've been asked to prove it via modular arithmetic. In order to do this I understand that I'm to use two lemmas:
- $\varphi(p^n) = p^{n} – p^{n-1}$. This I can prove by working out that there are $p^n$ numbers less than $p^n$ of which $p^{n-1}$ of them are divisible by $p$.
- Given $gcd(r,s) = 1$, $\varphi(rs) = \varphi(r)\varphi(s)$. This I'm having problems with.
My lecturer suggested the proposition that $\varphi(n) = n\prod_{p|n}\left(1-\frac{1}{p}\right)$. I can re-arrange this to equal lemma 1, but what I can't understand how that proposition proves 2, which my lecturer claims it does, and why these two points together prove the theorem.
I suspect I might follow the argument correctly if I fully understood lemma 2.
Best Answer
Here is another way to prove Euler's generalization. You do not need to know the formula of $\varphi(n)$ for this method which I think makes this method elegant.
Consider the set of all numbers less than $n$ and relatively prime to it. Let $\{a_1,a_2,...,a_{\varphi(n)}\}$ be this set.
Consider a number $c < n$ and relatively prime to it i.e. $c \in \{a_1,a_2,\ldots,a_{\varphi(n)}\}$.
First observe that for any $a_i$, $c a_{i} \equiv a_{j} \pmod{n}$ for some $j$. (True since $c$ and $a_i$ are themselves relatively prime to $n$, their product has to be relatively prime to $n$).
And if $c a_{i} \equiv c a_{j} \pmod{n}$ then $a_i = a_j$. (True as cancellation can be done since $c$ is relatively prime to $n$).
Hence, if we now consider the set $\{ca_1,ca_2,...,ca_{\varphi(n)}\}$ this is just a permutation of the set $\{a_1,a_2,...,a_{\varphi(n)}\}$.
Thereby, we have $\displaystyle \prod_{k=1}^{\varphi(n)} ca_k \equiv \prod_{k=1}^{\varphi(n)} a_k \pmod{n}$.
Hence, we get $\displaystyle c^{\varphi(n)} \prod_{k=1}^{\varphi(n)} a_k \equiv \prod_{k=1}^{\varphi(n)} a_k \pmod{n}$.
Now, note that $\displaystyle \prod_{k=1}^{\varphi(n)} a_k$ is relatively prime to $n$ and hence you can cancel them on both sides to get
$$c^{\varphi(n)} \equiv 1 \pmod{n}$$ whenever $(c,n) = 1$.