New, elegant proofs for $\varphi(p^{k})=p^{k}-p^{k-1}$

Question

Are there any short, elegant proofs known for the identity $\varphi(p^{k})=p^{k}-p^{k-1}$ ? (Here $\varphi$ is Euler's totient function and $p$ is a prime.)

The standard combinatorial proof goes like this:

In the set $\left\{ 1,2\ldots,p^{k}\right\} $ there in total $p^{k}$ number. Split this set into $p$ subsets $\left\{ 1,\ldots,p\right\} $, $\left\{ p+1,\ldots,2p\right\} \ldots$ Then in each of these sets there is only one number -- namely the one of the form $m\cdot p$ for some suitable $m$, that divides $p^{k}$. There are in total $\frac{p^{k}}{p}=p^{k-1}$ such sets, so in total $p^{k-1}$-many number from $\left\{ 1,2\ldots,p^{k}\right\} $ divide $p^{k}$. Thus $(p^{k}-p^{k-1})$-many numbers are coprime to $p^k$, which proves the identity. $\square$

(A different proof that is often encountered assumes that we know that $\varphi(n)=n\prod_{p\mid n}(1-\frac{1}{p})$, from which our identity follows immediately. But this is actually a longer proof, since proving the auxiliary identity is longer.)

Surprisingly, I would have imagined that there are tons of wildly different proofs of such a basic fact out there, but a preliminary internet seach as well as book skimming returned only (minor variations of) these two proofs.

EDIT The present proofs are more or less reformulations (very polished with details hidden as good as possible - but still reformulations) of my first proofs. What I'm looking for are more radically different approaches (if these exist).

Answer 1

A beautiful proof for the second identity you mention is probabilistic :

Take the set $\{1,...,n\}$ with the uniform probability measure (i.e. $P(\{i\}) = 1/n$ for any $i\in \{1,...,n\}$).

Then $\phi(n)/n = P(\{k, gcd(n,k) = 1\}) = P(\{k, \forall p$, prime, $(p$ divides $n) \implies (p$ doesn't divide $k)\} = P(\displaystyle\bigcap_{p\mid n} \{k, p$ doesn't divide $k\})$. Now it is easily proved that the events $(\{k, p$ divides $k\})_{p\mid n}$ are independent (just compute it) and so their complements are as well, which shows that

$\phi(n)/n = \prod_{p\mid n}(1- 1/p)$

That lets you conclude

Answer 2

Hint $\,\ \gcd(a,p^k)>1 \iff p\mid a \iff a\, \equiv\,\overbrace{ 1p,\,2p,3p,\ldots,\color{#c00}{p^{k-1}}p}^{\large\quad\color{#c00}{p^{\Large k-1}}\ \rm elements}\,\pmod{\!p^k}$

Thus there are $\,\color{#c00}{p^{k-1}}$ non-coprime residues, so $\,p^k - \color{#c00}{p^{k-1}}$ coprime residues mod $p^k$.

Answer 3

Yes, a one-line proof: $$(\mathbf Z/p^k\mathbf Z)^\times= \mathbf Z/p^k\mathbf Z\smallsetminus (p\mathbf Z/p^k\mathbf Z),\enspace\text{and}\quad p\mathbf Z/p^k\mathbf Z\simeq\mathbf Z/p^{k-1}\mathbf Z. $$ And a detail on a second line (well, a sesquiline…) for the isomorphism: \begin{align}p\mathbf Z/p^k\mathbf Z&\longrightarrow\mathbf Z/p^{k-1}\mathbf ,\\ px+p^k\mathbf Z&\longmapsto x+p^{k-1}\mathbf Z. \end{align}

Answer 4

Among the numbers in $\{1,2,\ldots,p^k\}$ exactly $p^{k-1}$ have at least one factor $p$. Since $p^k$ has no prime factors other than $p$ it follows that $\phi(p^k)=p^k-p^{k-1}$.

Answer 5

Denote by $C_r$ the cyclic group of order $r$ and by $g_r$ the number of generators of $C_r$. Then for a prime $p$ the homomorphism $C_{p^k} \rightarrow C_{p^{k-1}}: g \mapsto g^p$, for $k>1$, maps generators onto generators and the size of a preimage of an element by the homomorphism is $p$ so $g_{p^k} = pg_{p^{k-1}}$. for the case $k = 1$ obviously $g_p = p-1$, so by induction we have $g_{p^k} =p^{k-1}(p-1)$.

Answer 6

$\mathbf Z/n\mathbf Z$ is an additive group with $n$ elements and has $\varphi(d)$ elements of order $d$ for each divisor $d$ of n. This is a well known fact that leads to Gauss 'formula: $$\sum_{d|n} \varphi(d) = n$$ So $$p^{k+1} = \sum_{d|p^{k+1}} \varphi(d) = \varphi(p^{k+1}) + \sum_{d|p^k} \varphi(d) = \varphi(p^{k+1}) + p^k$$ and $$\varphi(p^{k+1}) = p^{k+1} - p^k = p^k (p-1)$$ QED

Answer 7

As the only prime that divides $p^k$ is $p$, we only need to look at integers in $\{1,\dots,p^k-1\}$ that have $p$ as a factor.

There are precisely $p^{k-1}-1$ of these.

So $\phi(p^k)=(p^k-1)-(p^{k-1}-1)=p^k-p^{k-1}$.

Answer 8

Curiously nobody has used the product form of the Euler totient's function. If you are still interested, here is a proof more on the analytical side of the theory.

Since

$$ \varphi(n) = n \prod_{p|n} \left( 1-\frac{1}{p} \right), $$

replace $n=p^\alpha$ and you are done. The proof of the formula you can find it here.