Exercise 4.2.3 (Number of solutions to $\sigma(x) = n$)

Proof. Some values of $\sigma$:

(a)

For $n>1$, $n$ and $1$ are distinct divisors of $n$, therefore $$\sigma (n)=\underset{d\mid n}{\sum }d\ge n+1.$$

If $n\ge 3$, $\sigma (n)\ge n+1\ge 4$, and $\sigma (1)=1,\sigma (2)=3$, thus $\sigma (x)=2$ has no solution. Since $1=\sigma (1)$, the smallest positive integer $n$ so that $\sigma (x)=n$ has no solution is

$$n=2.$$

(b)

First $\sigma (1)=1$. By part (a), if $n>1$, $\sigma (n)\ge n+1$, thus $\sigma (n)\ne 1$. The only solution of $\sigma (x)=1$ is $n=1$. So the smallest positive integer $n$ so that $\sigma (x)=n$ has exactly one solution is $$n=1.$$

(c)

We note that $\sigma (6)=\sigma (11)=12$ and the given array of values shows that the equation $\sigma (x)=12$ has no other solution less that $12$. If $n\ge 12$, $\sigma (n)\ge n+1>13$, so $\sigma (n)\ne 12$. Moreover, no value less that $12$ has two antecedents. This shows that the smallest positive integer $n$ so that $\sigma (x)=n$ has exactly two solutions is $$n=12.$$

(d)

We note that $\sigma (14)=\sigma (15)=\sigma (23)=24$ and the given array of values shows that the equation $\sigma (x)=24$ has no other solution less that $24$. If $n\ge 24$, $\sigma (n)\ge n+1>25$, so $\sigma (n)\ne 24$. Moreover, no value less that $24$ has three antecedents. This shows that the smallest positive integer $n$ so that $\sigma (x)=n$ has exactly three solutions is $$n=24.$$