# 4.2: Multiplicative Number Theoretic Functions

- Page ID
- 8839

We now present several multiplicative number theoretic functions which will play a crucial role in many number theoretic results. We start by discussing the Euler phi-function which was defined in an earlier chapter. We then define the sum-of-divisors function and the number-of-divisors function along with their properties.

## The Euler \(\phi\)-Function

As defined earlier, the Euler \(\phi\)-function counts the number of integers smaller than and relatively prime to a given integer. We first calculate the value of the \(phi\)-function at primes and prime powers.

If \(p\) is prime, then \(\phi(p)=p-1\). Conversely, if \(p\) is an integer such that \(\phi(p)=p-1\), then \(p\) is prime.

The first part is obvious since every positive integer less than \(p\) is relatively prime to \(p\). Conversely, suppose that \(p\) is not prime. Then \(p=1\) or \(p\) is a composite number. If \(p=1\), then \(\phi(p)\neq p-1\). Now if \(p\) is composite, then \(p\) has a positive divisor. Thus \(\phi(p)\neq p-1\). We have a contradiction and thus \(p\) is prime.

We now find the value of \(\phi\) at prime powers.

Let \(p\) be a prime and \(m\) a positive integer, then \(\phi(p^m)=p^m-p^{m-1}\).

Note that all integers that are relatively prime to \(p^m\) and that are less than \(p^m\) are those that are not multiple of \(p\). Those integers are \(p,2p,3p,...,p^{m-1}p\). There are \(p^{m-1}\) of those integers that are not relatively prime to \(p^m\) and that are less than \(p^m\). Thus \[\phi(p^m)=p^m-p^{m-1}.\]

\(\phi(7^3)=7^3-7^2=343-49=294\). Also \(\phi(2^{10})=2^{10}-2^9=512.\)

We now prove that \(\phi\) is a multiplicative function.

Let \(m\) and \(n\) be two relatively prime positive integers. Then \(\phi(mn)=\phi(m)\phi(n)\).

Denote \(\phi(m)\) by \(s\) and let \(k_1,k_2,...,k_s\) be a reduced residue system modulo \(m\). Similarly, denote \(\phi(n)\) by \(t\) and let \(k_1',k_2',...,k_t'\) be a reduced residue system modulo \(n\). Notice that if \(x\) belongs to a reduced residue system modulo \(mn\), then \[(x,m)=(x,n)=1.\] Thus \[x\equiv k_i(mod\ m) \mbox{and} \ \ x\equiv k_j'(mod \ n)\] for some \(i,j\). Conversely, if \[x\equiv k_i(mod\ m) \mbox{and} \ \ x\equiv k_j'(mod \ n)\] some \(i,j\) then \((x,mn)=1\) and thus \(x\) belongs to a reduced residue system modulo \(mn\). Thus a reduced residue system modulo \(mn\) can be obtained by by determining all \(x\) that are congruent to \(k_i\) and \(k_j'\) modulo \(m\) and \(n\) respectively. By the Chinese remainder theorem, the system of equations \[x\equiv k_i(mod\ m) \mbox{and} \ \ x\equiv k_j'(mod \ n)\] has a unique solution. Thus different \(i\) and \(j\) will yield different answers. Thus \(\phi(mn)=st\).

We now derive a formula for \(\phi(n)\).

Let \(n=p_1^{a_1}p_2^{a_2}...p_s^{a_s}\) be the prime factorization of \(n\). Then \[\phi(n)=n\left(1-\frac{1}{p_1}\right)\left(1-\frac{1}{p_2}\right)...\left(1-\frac{1}{p_s}\right).\]

By Theorem 37, we can see that for all \(1\leq i\leq k\) \[\phi(p_i^{a_i})=p_i^{a_i}-p_i^{a_i-1}=p_i^{a_i}\left(1-\frac{1}{p_i}\right).\] Thus by Theorem 38, \[\begin{aligned} \phi(n)&=&\phi(p_1^{a_1}p_2^{a_2}...p_s^{a_s})\\&=& \phi(p_1^{a_1})\phi(p_2^{a_2})...\phi(p_s^{a_s})\\&=&p_1^{a_1}\left(1-\frac{1}{p_1}\right) p_2^{a_2}\left(1-\frac{1}{p_2}\right)...p_s^{a_s}\left(1-\frac{1}{p_s}\right)\\&=& p_1^{a_1}p_2^{a_2}...p_k^{a_k}\left(1-\frac{1}{p_1}\right)\left(1-\frac{1}{p_2}\right)... \left(1-\frac{1}{p_s}\right)\\&=& n\left(1-\frac{1}{p_1}\right)\left(1-\frac{1}{p_2}\right)...\left(1-\frac{1}{p_s}\right).\end{aligned}\]

Note that \[\phi(200)=\phi(2^35^2)=200\left(1-\frac{1}{2}\right)\left(1-\frac{1}{5}\right)=80.\]

Let \(n\) be a positive integer greater than 2. Then \(\phi(n)\) is even.

Let \(n=p_1^{a_1}p_2^{a_2}...p_k^{a_k}\). Since \(\phi\) is multiplicative, then \[\phi(n)=\prod_{j=1}^k\phi(p_j^{a_j}).\] Thus by Theorem 39, we have \[\phi(p_j^{a_j})=p_j^{a_j-1-1}(p_j-1).\] We see then \(\phi(p_j^{a_j})\)is even if \(p_j\) is an odd prime. Notice also that if \(p_j=2\), then it follows that \(\phi(p_j^{a_j})\) is even. Hence \(\phi(n)\) is even.

Let \(n\) be a positive integer. Then \[\sum_{d\mid n}\phi(d)=n.\]

Split the integers from 1 to \(n\) into classes. Put an integer \(m\) in the class \(C_d\) if the greatest common divisor of \(m\) and \(n\) is \(d\). Thus the number of integers in the \(C_d\) class is the number of positive integers not exceeding \(n/d\) that are relatively prime to n/d. Thus we have \(\phi(n/d)\) integers in \(C_d\). Thus we see that \[n=\sum_{d\mid n}\phi(n/d).\] As \(d\) runs over all divisors of \(n\), so does \(n/d\). Hence \[n=\sum_{d\mid n}\phi(n/d)=\sum_{d\mid n}\phi(d).\]

## The Sum-of-Divisors Function

The sum of divisors function, denoted by \(\sigma(n)\), is the sum of all positive divisors of \(n\).

\(\sigma(12)=1+2+3+4+6+12=28.\)

Note that we can express \(\sigma(n)\) as \(\sigma(n)=\sum_{d\mid n}d\).

We now prove that \(\sigma(n)\) is a multiplicative function.

The sum of divisors function \(\sigma(n)\) is multiplicative.

We have proved in Theorem 35 that the summatory function is multiplicative once \(f\) is multiplicative. Thus let \(f(n)=n\) and notice that \(f(n)\) is multiplicative. As a result, \(\sigma(n)\) is multiplicative.

Once we found out that \(\sigma(n)\) is multiplicative, it remains to evaluate \(\sigma(n)\) at powers of primes and hence we can derive a formula for its values at any positive integer.

Let \(p\) be a prime and let \(n=p_1^{a_1}p_2^{a_2}...p_t^{a_t}\) be a positive integer. Then \[\sigma(p^a)=\frac{p^{a+1}-1}{p-1},\] and as a result, \[\sigma(n)=\prod_{j=1}^{t}\frac{p_j^{a_j+1}-1}{p_j-1}\]

Notice that the divisors of \(p^{a}\) are \(1,p,p^2,...,p^a\). Thus \[\sigma(p^a)=1+p+p^2+...+p^a=\frac{p^{a+1}-1}{p-1}.\] where the above sum is the sum of the terms of a geometric progression.

Now since \(\sigma(n)\) is multiplicative, we have \[\begin{aligned} \sigma(n)&=&\sigma(p^{a_1})\sigma(p^{a_2})...\sigma(p^{a_t})\\&=& \frac{p_1^{a_1+1}-1}{p_1-1}.\frac{p_2^{a_2+1}-1}{p_2-1}...\frac{p_t^{a_t+1}-1}{p_t-1}\\ &=&\prod_{j=1}^{t}\frac{p_j^{a_j+1}-1}{p_j-1}\end{aligned}\]

\(\sigma(200)=\sigma(2^35^2)=\frac{2^4-1}{2-1}\frac{5^3-1}{5-1}=15.31=465.\)

## The Number-of-Divisors Function

The number of divisors function, denoted by \(\tau(n)\), is the sum of all positive divisors of \(n\).

\(\tau(8)=4.\)

We can also express \(\tau(n)\) as \(\tau(n)=\sum_{d\mid n}1\).

We can also prove that \(\tau(n)\) is a multiplicative function.

The number of divisors function \(\tau(n)\) is multiplicative.

By Theorem 36, with \(f(n)=1\), \(\tau(n)\) is multiplicative.

We also find a formula that evaluates \(\tau(n)\) for any integer \(n\).

Let \(p\) be a prime and let \(n=p_1^{a_1}p_2^{a_2}...p_t^{a_t}\) be a positive integer. Then \[\tau(p^a)=a+1,\] and as a result, \[\tau(n)=\prod_{j=1}^{t}(a_j+1).\]

The divisors of \(p^{a}\) as mentioned before are \(1,p,p^2,...,p^a\). Thus \[\tau(p^a)=a+1\]

Now since \(\tau(n)\) is multiplicative, we have \[\begin{aligned} \tau(n)&=&\tau(p^{a_1})\tau(p^{a_2})...\tau(p^{a_t})\\&=& (a_1+1)(a_2+1)...(a_t+1)\\ &=&\prod_{j=1}^{t}(a_j+1).\end{aligned}\]

\(\tau(200)=\tau(2^35^2)=(3+1)(2+1)=12\).

**Exercises**

- Find \(\phi(256)\) and \(\phi(2.3.5.7.11)\).
- Show that \(\phi(5186)=\phi(5187)\).
- Find all positive integers \(n\) such that \(\phi(n)=6\).
- Show that if \(n\) is a positive integer, then \(\phi(2n)=\phi(n)\) if \(n\) is odd.
- Show that if \(n\) is a positive integer, then \(\phi(2n)=2\phi(n)\) if \(n\) is even.
- Show that if \(n\) is an odd integer, then \(\phi(4n)=2\phi(n)\).
- Find the sum of positive integer divisors and the number of positive integer divisors of 35
- Find the sum of positive integer divisors and the number of positive integer divisors of \(2^53^45^37^313\).
- Which positive integers have an odd number of positive divisors.
- Which positive integers have exactly two positive divisors.

## Contributors and Attributions

Dr. Wissam Raji, Ph.D., of the American University in Beirut. His work was selected by the Saylor Foundation’s

for public release under a Creative Commons Attribution (**Open Textbook Challenge**) license.**CC BY**