Perfect Number has at least Two Distinct Prime Factors
Theorem
Let $n \in \N$ be a perfect number.
Then $n$ has at least two distinct prime factors.
Proof 1
Aiming for a contradiction, suppose the contrary: that $n$ is a perfect number with exactly $1$ prime factor.
Hence let $n = p^k$ where $p$ is prime.
By definition of perfect number:
- $\map {\sigma_1} n = 2 n$
where $\map {\sigma_1} n$ denotes the divisor sum of $n$.
Hence:
\(\ds \map {\sigma_1} n\) | \(=\) | \(\ds 2 n\) | Definition of Perfect Number | |||||||||||
\(\ds \map {\sigma_1} {p^k}\) | \(=\) | \(\ds \dfrac {p^{k + 1} - 1} {p - 1}\) | Divisor Sum of Power of Prime | |||||||||||
\(\ds \) | \(=\) | \(\ds 1 + p + \cdots + p^k\) | Sum of Geometric Sequence | |||||||||||
\(\ds \) | \(=\) | \(\ds 2 p^k\) | Definition of $n$ | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds 1\) | \(=\) | \(\ds 2 p^k - \paren {p + \cdots + p^k}\) | |||||||||||
\(\ds \) | \(=\) | \(\ds p \paren {2 p^{k - 1} - \paren {1 + p + \cdots + p^{k - 1} } }\) | simplification | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds p\) | \(\divides\) | \(\ds 1\) |
That is, $p$ is a divisor of $1$ which is a contradiction.
This article, or a section of it, needs explaining. In particular: To be rigorous, we need to specify explicitly what it contradicts: that $p > 1$ but $p \divides 1 \implies p \le 1$ You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Explain}} from the code. |
Hence the result by Proof by Contradiction.
$\blacksquare$
Proof 2
Aiming for a contradiction, suppose the contrary: that $n$ is a perfect number with exactly $1$ prime factor.
Hence let $n = p^k$ where $p$ is prime.
By definition of perfect number:
- $\map {\sigma_1} n = 2 n$
where $\map {\sigma_1} n$ denotes the divisor sum of $n$.
Hence:
\(\ds \map {\sigma_1} n\) | \(=\) | \(\ds 2 n\) | Definition of Perfect Number | |||||||||||
\(\ds \map {\sigma_1} {p^k}\) | \(=\) | \(\ds \dfrac {p^{k + 1} - 1} {p - 1}\) | Divisor Sum of Power of Prime | |||||||||||
\(\ds \) | \(=\) | \(\ds 1 + p + \cdots + p^k\) | Sum of Geometric Sequence | |||||||||||
\(\ds \) | \(=\) | \(\ds 2 p^k\) | Definition of $n$ | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac {\map {\sigma_1} {p^k} } p^k\) | \(=\) | \(\ds 2\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map h n\) | \(=\) | \(\ds 2\) | where $\map h n$ denotes the abundancy index of $n$ | ||||||||||
\(\ds \) | \(<\) | \(\ds \dfrac p {p - 1}\) | Upper Bound for Abundancy Index | |||||||||||
\(\ds \) | \(<\) | \(\ds \dfrac 3 2\) | as $\dfrac x {x - 1}$ is decreasing on $x > 1$ | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds 2\) | \(<\) | \(\ds \dfrac 3 2\) | which contradicts the fact that $2 > \dfrac 3 2$ |
Hence the result by Proof by Contradiction.
$\blacksquare$