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

## Theorem

Let:

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

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:

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

The divisors of these numbers are as follows:

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

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

Hence the result.

$\blacksquare$