The Number of Positive Divisors of an Integer Examples 2

# The Number of Positive Divisors of an Integer Examples 2

Recall from The Number of Positive Divisors of an Integer page that if $n \in \mathbb{Z}$ then the number of positive divisors of $n$ is denoted $d(n)$ and is given explicitly by:

(1)
\begin{align} \quad d(n) = \sum_{d \mid n}_{d > 0} 1 \end{align}

Most importantly, we noted that:

• If $p$ is prime then $d(p) = 2$.
• If $p$ is prime and $k \in \mathbb{N}$ then $d(p^k) = k + 1$.
• If $p$ and $q$ are prime then $d(pq) = d(p)d(q)$.

We will now look at some examples of computing regarding the number of positive divisors of an integer. More examples can be found on The Number of Positive Divisors of an Integer Examples 1 page.

## Example 1

For which primes $p$ is $p \cdot d(p) ≥ p^2$ ?

We must first note that $d(p) = 2$ for all primes $p$. Hence we need to find primes p such that $2p ≥ p^2$. We note that this is only true when $p = 2$ since $4 ≥ 4$. For all other primes, $2p < p^2$. Hence $p = 2$ is the only solution to this problem.

## Example 2

For which primes $p$ and $q$ does $pq \cdot d(pq) ≥ p^2q^2$?

Since p and q are both primes, it follows that $(p, q) = 1$, and hence $d(pq) = d(p)d(q) = (2)(2) = 4$. Hence we want to find primes $p$ and $q$ that are true for the inequality $4pq ≥ p^2 q^2$.

First, fix $p = 2$. Hence we want to solve the inequality that $8q ≥ 4q^2$, or more appropriately $2q ≥ q^2$. We've already solved this in example 6 though! Hence q = 2 is our only solution. So if p = 2 and q = 2, then $4pq ≥ p^2q^2$.

Now fix $p = 3$. Hence we now want to solve the inequality that $12q ≥ 9q^2$, or rather $4q ≥ 3q^2$. But if $q$ is prime, then this is never true. In fact, $4pq ≥ p^2 q^2$ is ONLY true if $p = q = 2$. Hence we have only one solution.

## Example 3

For which primes $p$ and $q$ does $d(p^q) > d(q^p)$.

We note that since $p$ and $q$ are both primes that $d(p^q) = q + 1$ and $d(q^p) = p + 1$. Hence we want to find values of $p$ and $q$ that satisfy $q + 1 > p + 1$, which reduces down to $q > p$. Hence for all primes q > p, $d(p^q) > d(p^q)$.

## Example 4

Show that $d(n)$ is odd if and only if $n$ is a square.

First, suppose that $d(n)$ is odd. Since the positive divisors of $n$ come in pairs, this must mean that there exists a positive divisor $d$ that is repeated. Thus $n = d^2$ for some $d \in \mathbb{N}$.

Conversely, if $n$ is a square, then $n = d^2$ for some $d \in \mathbb{N}$. Suppose the prime power decomposition of $d = p_1^{e_1}p_2^{e_2}...p_k^{e_k}$. It thus follows that $n = p_1^{2e_1}p_2^{2e_2}...p_k^{2e_k}$. Hence:

(2)
\begin{equation} d(n) = (2e_1 + 1)(2e_2 + 1)...(2e_k + 1) \end{equation}

Since all $2e_i + 1$ are odd, it thus follows that the product will also be odd, and hence $d(n)$ is odd too.