Exercise 4.2.14 (There is only a finite number of integers $x$ satisfying $\sigma(x) = n$)

Question

Given any positive integer $n$, prove that there is only a finite number of integers $x$ satisfying $\sigma(x) = n$.

Richard Ganaye · Accepted Answer

ewcommand{\N}{\mathbb{N}}

\begin{proof} 
If $x >1$, $1$ and $x$ are distinct divisors of $x$, thus
$$\sigma(x) = \sum_{d \mid x} d \geq x + 1\qquad (x>1).$$
A fortiori, for all positive integer $x$,
$$\sigma(x) \geq x\qquad (x  \geq 1).$$
Let $n \geq 1$. Then every $x$ such that $\sigma(x) = n$ satisfies $x \leq n$ (otherwise $\sigma(x) \geq x >n$, and so $\sigma(x) 
e n$). If 
$$A = \{x \in \N^* \mid \sigma(x) = n\},$$
then $A \subset [\![1,n]\!]$, so $A$ is a finite set.

There is only a finite number of integers $x$ satisfying $\sigma(x) = n$.
\end{proof}

Exercise 4.2.14 (There is only a finite number of integers $x$ satisfying $\sigma(x) = n$)

Answers

Comments

Exercise 4.2.14 (There is only a finite number of integers $x$ satisfying $\sigma(x) = n$)

Answers

Comments

Add answer