Perfect Number has at least Two Distinct Prime Factors/Proof 1

Theorem

Let $n \in \N$ be a perfect number.

Then $n$ has at least two distinct prime factors.

Proof

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.

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Hence the result by Proof by Contradiction.

$\blacksquare$