Famous Theorems of Mathematics/Number Theory/Totient Function

This page provides proofs for identities involving the totient function $$\varphi(k)$$ and the Möbius function $$\mu(k)$$.

Sum of integers relatively prime to and less than or equal to n
The proof of the identity


 * $$\sum_{1\le k\le n \atop (k,n)=1} k = \frac{1}{2} \, \varphi(n) \, n

\quad \mbox{ where } \quad n \ge 2$$

uses the fact that


 * $$ (k, n) = d \quad \Leftrightarrow \quad (n-k, n) = d $$

because if $$d|n$$ and $$d|k$$ then $$d|n-k$$ and if $$d|n$$ and $$d|(n-k)$$ then $$d|n-(n-k) = k.$$

This means that for $$n>2$$ we may group the k that are relatively prime to n into pairs
 * $$(k, n-k) \quad \mbox{ where } \quad k1$$ because we assumed that $$n>2.$$ There are
 * $$\frac{\varphi(n)}{2}$$

such pairs, and the constituents of each pair sum to
 * $$k \; + \; n-k \; = n,$$

hence
 * $$\sum_{(k, n)=1} k = \frac{1}{2} \, \varphi(n) \, n

\quad \mbox{ where } \quad n \ge 3.$$ The case $$n=2$$ is verified by direct substitution and may be included in the formula.

Proofs of totient identities involving the floor function
The proof of the identity


 * $$\sum_{k=1}^n\frac{\varphi(k)}{k} = \sum_{k=1}^n\frac{\mu(k)}{k}\left\lfloor\frac{n}{k}\right\rfloor$$

is by mathematical induction on $$n$$. The base case is $$n=1$$ and we see that the claim holds:


 * $$\varphi(1)/1 = 1 = \frac{\mu(1)}{1} \left\lfloor 1\right\rfloor.$$

For the induction step we need to prove that


 * $$\frac{\varphi(n+1)}{n+1} =

\sum_{k=1}^n\frac{\mu(k)}{k} \left(\left\lfloor\frac{n+1}{k}\right\rfloor - \left\lfloor\frac{n}{k}\right\rfloor\right) + \frac{\mu(n+1)}{n+1}.$$

The key observation is that



\left\lfloor\frac{n+1}{k}\right\rfloor - \left\lfloor\frac{n}{k}\right\rfloor \; = \; \begin{cases} 1, & \mbox{if }k|(n+1) \\ 0, & \mbox{otherwise, } \end{cases}$$

so that the sum is



\sum_{k|n+1,\; k<n+1} \frac{\mu(k)}{k} + \frac{\mu(n+1)}{n+1} = \sum_{k|n+1} \frac{\mu(k)}{k}.$$

Now the fact that



\sum_{k|n+1} \frac{\mu(k)}{k} = \frac{\varphi(n+1)}{n+1} $$

is a basic totient identity. To see that it holds, let $$p_1^{v_1} p_2^{v_2} \ldots p_q^{v_q}$$ be the prime factorization of n+1. Then


 * $$\frac{\varphi(n+1)}{n+1} = \prod_{l=1}^q \left( 1 - \frac{1}{p_l} \right)

= \sum_{k|n+1} \frac{\mu(k)}{k} $$

by definition of $$\mu(k).$$ This concludes the proof.

An alternate proof proceeds by substituting $$\frac{\varphi(k)}{k} = \sum_{d|k}\frac{\mu(d)}{d}$$ directly into the left side of the identity, giving $$\sum_{k=1}^n \sum_{d|k}\frac{\mu(d)}{d}.$$

Now we ask how often the term $$\begin{matrix}\frac{\mu(d)}{d}\end{matrix}$$ occurs in the double sum. The answer is that it occurs for every multiple $$k$$ of $$d$$, but there are precisely $$\begin{matrix}\left\lfloor\frac{n}{d}\right\rfloor\end{matrix}$$ such multiples, which means that the sum is


 * $$\sum_{d=1}^n\frac{\mu(d)}{d}\left\lfloor\frac{n}{d}\right\rfloor$$

as claimed.

The trick where zero values of $$\begin{matrix} \left\lfloor\frac{n+1}{k}\right\rfloor - \left\lfloor\frac{n}{k}\right\rfloor \end{matrix}$$ are filtered out may also be used to prove the identity


 * $$\sum_{k=1}^n\varphi(k) = \frac{1}{2}\left(1+ \sum_{k=1}^n \mu(k)\left\lfloor\frac{n}{k}\right\rfloor^2\right).$$

The base case is $$n=1$$ and we have


 * $$\varphi(1) = 1 =

\frac{1}{2} \left(1+ \mu(1) \left\lfloor\frac{1}{1}\right\rfloor^2\right)$$

and it holds. The induction step requires us to show that


 * $$\varphi(n+1) = \frac{1}{2}

\sum_{k=1}^n \mu(k) \left( \left\lfloor\frac{n+1}{k}\right\rfloor^2 - \left\lfloor\frac{n}{k}\right\rfloor^2 \right) \; + \; \frac{1}{2} \; \mu(n+1) \; \left\lfloor\frac{n+1}{n+1}\right\rfloor^2 .$$

Next observe that



\left\lfloor\frac{n+1}{k}\right\rfloor^2 - \left\lfloor\frac{n}{k}\right\rfloor^2 \; = \; \begin{cases} 2\frac{n+1}{k} - 1, & \mbox{if }k|(n+1) \\ 0, & \mbox{otherwise.} \end{cases}$$

This gives the following for the sum

\frac{1}{2} \sum_{k|n+1, \; k<n+1} \mu(k) \left( 2\frac{n+1}{k} - 1 \right) \; + \; \frac{1}{2} \; \mu(n+1) = \frac{1}{2} \sum_{k|n+1} \mu(k) \left( 2 \; \frac{n+1}{k} - 1 \right). $$

Treating the two inner terms separately, we get



(n+1) \sum_{k|n+1} \frac{\mu(k)}{k} - \frac{1}{2} \sum_{k|n+1} \mu(k). $$

The first of these two is precisely $$\varphi(n+1)$$ as we saw earlier, and the second is zero, by a basic property of the Möbius function (using the same factorization of $$n+1$$ as above, we have $$\begin{matrix} \sum_{k|n+1} \mu(k) = \prod_{l=1}^q (1-1) = 0 \end{matrix}$$.) This concludes the proof.

This result may also be proved by inclusion-exclusion. Rewrite the identity as
 * $$-1 + 2\sum_{k=1}^n\varphi(k) =

\sum_{k=1}^n \mu(k)\left\lfloor\frac{n}{k}\right\rfloor^2.$$

Now we see that the left side counts the number of lattice points (a, b) in [1,n] &times; [1,n] where a and b are relatively prime to each other. Using the sets $$A_p$$ where p is a prime less than or equal to n to denote the set of points where both coordinates are divisible by p we have


 * $$ \left| \bigcup_p A_p \right| =

\sum_p \left| A_p \right| \; - \; \sum_{p<q} \left| A_p \cap A_q \right| \; + \; \sum_{p<q<r} \left| A_p \cap A_q \cap A_r \right| \; - \; \cdots \; \pm \; \left| A_p \cap \; \cdots \; \cap A_s \right|. $$

This formula counts the number of pairs where a and b are not relatively prime to each other. The cardinalities are as follows:



\left| A_p \right| = \left\lfloor \frac{n}{p} \right\rfloor^2, \; \left| A_p \cap A_q \right| = \left\lfloor \frac{n}{pq} \right\rfloor^2, \; \left| A_p \cap A_q \cap A_r \right| = \left\lfloor \frac{n}{pqr} \right\rfloor^2, \; \ldots $$

and the signs are $$\begin{matrix}-\mu(pqr\cdots)\end{matrix}$$, hence the number of points with relatively prime coordinates is

$$\mu(1)\, n^2 \; + \; \sum_p \mu(p) \left\lfloor \frac{n}{p} \right\rfloor^2 \; + \; \sum_{p<q} \mu(p q) \left\lfloor \frac{n}{pq} \right\rfloor^2 \; + \; \sum_{p<q<r} \mu(p q r) \left\lfloor \frac{n}{pqr} \right\rfloor^2 \; + \; \cdots $$

but this is precisely $$\sum_{k=1}^n \mu(k) \left\lfloor \frac{n}{k} \right\rfloor^2$$ and we have the claim.

Average order of the totient
We will use the last formula of the preceding section to prove the following result:
 * $$\frac{1}{n^2} \sum_{k=1}^n \varphi(k)=

\frac{3}{\pi^2} + \mathcal{O}\left(\frac{\log n }{n}\right)$$

Using $$x-1 < \lfloor x \rfloor \le x$$ we have the upper bound
 * $$\frac{1}{2 n^2}

\left(1+ \sum_{k=1}^n \mu(k)\frac{n^2}{k^2}\right) = \frac{1}{2 n^2} + \frac{1}{2} \sum_{k=1}^n \frac{\mu(k)}{k^2} $$

and the lower bound
 * $$\frac{1}{2 n^2}

\left(1+ \sum_{k=1}^n \mu(k)\left(\frac{n^2}{k^2} - 2\frac{n}{k} + 1\right)\right)$$ which is

\frac{1}{2 n^2} + \frac{1}{2} \sum_{k=1}^n \frac{\mu(k)}{k^2} - \frac{1}{n} \sum_{k=1}^n \frac{\mu(k)}{k} + \frac{1}{2 n^2} \sum_{k=1}^n \mu(k) $$

Working with the last two terms and using the asymptotic expansion of the nth harmonic number we have
 * $$- \frac{1}{n} \sum_{k=1}^n \frac{\mu(k)}{k} >

- \frac{1}{n} \sum_{k=1}^n \frac{1}{k} = - \frac{1}{n} H_n > -\frac{1}{n} (\log n +1) $$ and
 * $$\frac{1}{2 n^2} \sum_{k=1}^n \mu(k) > - \frac{1}{2 n}.$$

Now we check the order of the terms in the upper and lower bound. The term $$\begin{matrix}\sum_{k=1}^n \frac{\mu(k)}{k^2}\end{matrix}$$ is $$\mathcal{O}(1)$$ by comparison with $$\zeta(2)$$, where $$\zeta(s)$$ is the Riemann zeta function. The next largest term is the logarithmic term from the lower bound.

So far we have shown that
 * $$\frac{1}{n^2} \sum_{k=1}^n \varphi(k)=

\frac{1}{2} \sum_{k=1}^n \frac{\mu(k)}{k^2} + \mathcal{O}\left(\frac{\log n }{n}\right)$$

It remains to evaluate $$\begin{matrix}\sum_{k=1}^n \frac{\mu(k)}{k^2}\end{matrix}$$ asymptotically, which we have seen converges. The Euler product for the Riemann zeta function is
 * $$\zeta(s) = \prod_p \left(1 - \frac{1}{p^s} \right)^{-1}

\mbox{ for } \Re(s)>1. $$

Now it follows immediately from the definition of the Möbius function that
 * $$\frac{1}{\zeta(s)} = \prod_p \left(1 - \frac{1}{p^s} \right)

= \sum_{n \ge 1} \frac{\mu(n)}{n^s}.$$

This means that
 * $$\frac{1}{2} \sum_{k=1}^n \frac{\mu(k)}{k^2} =

\frac{1}{2} \frac{1}{\zeta(2)} + \mathcal{O}\left(\frac{1}{n}\right)$$ where the integral $$\begin{matrix} \int_{n+1}^\infty \frac{1}{t^2} dt\end{matrix}$$ was used to estimate $$\begin{matrix} \sum_{k>n} \frac{\mu(k)}{k^2}\end{matrix}.$$ But $$\begin{matrix}\frac{1}{2} \frac{1}{\zeta(2)} = \frac{3}{\pi^2}\end{matrix}$$ and we have established the claim.

Average order of &phi;(n)/n
The material of the preceding section, together with the identity


 * $$\sum_{k=1}^n\frac{\varphi(k)}{k} = \sum_{k=1}^n\frac{\mu(k)}{k}\left\lfloor\frac{n}{k}\right\rfloor$$

also yields a proof that


 * $$\frac{1}{n} \sum_{k=1}^n \frac{\varphi(k)}{k} =

\frac{6}{\pi^2} + \mathcal{O}\left(\frac{\log n }{n}\right).$$

Reasoning as before, we have the upper bound
 * $$\frac{1}{n} \sum_{k=1}^n\frac{\mu(k)}{k}\frac{n}{k} =

\sum_{k=1}^n \frac{\mu(k)}{k^2}$$ and the lower bound
 * $$-\frac{1}{n} \sum_{k=1}^n\frac{\mu(k)}{k} +

\sum_{k=1}^{n+1} \frac{\mu(k)}{k^2}.$$

Now apply the estimates from the preceding section to obtain the result.

Inequalities
We first show that


 * $$\lim \inf \frac{\varphi (n)}{n}=0 \mbox{ and } \lim \sup \frac{\varphi (n)}{n}=1. $$

The latter holds because when n is a power of a prime p, we have


 * $$\frac{\varphi (n)}{n} = 1 - \frac{1}{p},$$

which gets arbitrarily close to 1 for p large enough (and we can take p as large as we please since there are infinitely many primes).

To see the former, let nk be the product of the first k primes, call them $$p_1, p_2, ..., p_k$$. Let


 * $$r_k = \frac{\varphi (n_k)}{n_k} = \prod_{i=1}^k \left( 1 - \frac{1}{p_i} \right).$$

Then


 * $$\frac{1}{r_k} = \prod_{i=1}^k \left( 1 - \frac{1}{p_i} \right)^{-1} >

\sum_{m=1}^{p_k} \frac{1}{m} = H_{p_k},$$

a harmonic number. Hence, by the well-known bound $$H_n > \log n$$, we have


 * $$\frac{1}{r_k} > \log p_k.$$

Since the logarithm is unbounded, taking k arbitrarily large ensures that rk achieves values arbitrarily close to zero.

We use the same factorization of n as in the first section to prove that


 * $$\frac {6 n^2} {\pi^2} < \varphi(n) \sigma(n) < n^2$$.

Note that


 * $$ \varphi(n) \sigma(n) =

n \prod_{l=1}^q \left(1 - \frac{1}{p_l}\right) \prod_{l=1}^q \frac{p_l^{v_l+1}-1}{p_l-1} = n \prod_{l=1}^q \frac{p_l-1}{p_l} \; \frac{p_l^{v_l+1}-1}{p_l-1}$$

which is



n \prod_{l=1}^q \left(p_l^{v_l}-\frac{1}{p_l}\right) = n^2 \prod_{l=1}^q \left(1 - \frac{1}{p_l^{v_l+1}}\right).$$

The upper bound follows immediately since


 * $$\prod_{l=1}^q \left(1 - \frac{1}{p_l^{v_l+1}}\right) < 1.$$

We come arbitrarily close to this bound when n is prime. For the lower bound, note that


 * $$ \prod_{l=1}^q \left(1 - \frac{1}{p_l^{v_l+1}}\right) \ge

\prod_{l=1}^q \left(1 - \frac{1}{p_l^2}\right) > \prod_p \left(1 - \frac{1}{p^2}\right),$$

where the product is over all primes. We have already seen this product, as in


 * $$ \prod_p \left(1 - \frac{1}{p^s}\right) =

\sum_{n\ge 1} \frac{\mu(n)}{n^s} = \frac{1}{\zeta(s)}$$

so that


 * $$\prod_p \left(1 - \frac{1}{p^2}\right) = \frac{1}{\zeta(2)}

= \frac{6}{\pi^2}$$

and we have the claim. The values of n that come closest to this bound are products of the first k primes.