# Factorization of Natural Numbers within 4 n + 1 not Unique

## Theorem

Let:

$S = \set {4 n + 1: n \in \N} = \set {1, 5, 9, 13, 17, \ldots}$

be the set of natural numbers of the form $4 n + 1$.

Then not all elements of $S$ have a complete factorization by other elements of $S$ which is unique.

## Proof

Consider the number:

$m = 693 = 3^2 \times 7 \times 11$

Thus:

$m = 9 \times 77 = 21 \times 33$

We have that:

 $\ds 9$ $=$ $\ds 4 \times 2 + 1$ $\ds \in S$ $\ds 77$ $=$ $\ds 4 \times 19 + 1$ $\ds \in S$ $\ds 21$ $=$ $\ds 4 \times 5 + 1$ $\ds \in S$ $\ds 33$ $=$ $\ds 4 \times 8 + 1$ $\ds \in S$

The divisors of these numbers are as follows:

 $\ds 9$ $=$ $\ds 3^2$ where $3 \notin S$ $\ds 77$ $=$ $\ds 7 \times 11$ where $7 \notin S$ and $11 \notin S$ $\ds 21$ $=$ $\ds 3 \times 7$ where $3 \notin S$ and $7 \notin S$ $\ds 33$ $=$ $\ds 3 \times 11$ where $3 \notin S$ and $11 \notin S$

Thus $693$ has two different complete factorizations into elements of $S$.

Hence the result.

$\blacksquare$