A Fundamental Study of Composite Numbers as a Different Perspective on Problems Related to Prime Numbers

1. Introduction

Studying prime numbers can be crucial because, among intelligent beings, prime numbers will always have a distinct and unmistakable meaning, regardless of the choice of counting basis and independent of mathematical notation. The riddle surrounding prime numbers, specifically their distribution, has eluded many bright and inquisitive researchers despite their best efforts [1]. So, the mystery surrounding the prime numbers continues to grab people’s attention.

Composite numbers can be defined as non-prime numbers when evaluated within integers other than 0. So, of course, it is possible to say that there is an impressive interplay between prime numbers and composite numbers and to find incredible correlations between prime numbers and composite numbers.

In this study, basic formulas for composite numbers have been developed so that composite numbers can be used frequently in prime numbers interpretation studies.

At the beginning of the study, it will be useful to give a few definitions, and more:

Preliminary Information I: Book VII - proposition 30 of Euclid’s Elements is the key in the proof of the fundamental theorem of arithmetic [2].

Preliminary Information II:Let x and y be any integer. According to the definition of even numbers, every even number is expressed as $2x$, and according to the definition of odd numbers, every odd number is expressed as $2y+1$.

Def. I: Prime numbers are numbers greater than 1 that do not have a factor other than themselves and 1 by the fundamental theorem of arithmetic [3,4].

Def. II: Composite numbers are numbers greater than 1 that do have a factor other than themselves and 1 by the fundamental theorem of arithmetic [3,5].

Basis I: Each composite number is expressed as the unique product of more than one prime number by the fundamental theorem of arithmetic and Book VII - proposition 31 & Book IX - proposition 14 of Euclid’s Elements [2,6].

Basis II: Every positive integer is either a prime number or a composite number by Def. I & II, Basis I and Book VII - proposition 32 of Euclid’s Elements [2].

NOTATIONS

$\mathbb{P}$	the set of all prime numbers
$\mathbb{C}$	the set of all composite numbers
$\mathbb{E}$	the set of all even numbers
$\mathbb{O}$	the set of all odd numbers
${\mathbb{N}}^{+}$	the set of all positive natural numbers
$a\in \mathbb{K}$	the element sign: a is an $element$ of the set $\mathbb{K}$
$p\wedge q$	the and sign: A conjunction of propositions read as “p $and$ q”

Table 1. Truth table of conjunction

p	q	$\mathit{p}\wedge \mathit{q}$
1	1	1
1	0	0
0	1	0
0	0	0

Table 1. Truth table of conjunction

p	q	$\mathit{p}\wedge \mathit{q}$
1	1	1
1	0	0
0	1	0
0	0	0

$p\oplus q$

the xor sign: An exclusive disjunction of propositions read as “p or $\left(xor\right)$ q”

Table 2. Truth table of exclusive or

p	q	$\mathit{p}\oplus \mathit{q}$
1	1	0
1	0	1
0	1	1
0	0	0

Table 2. Truth table of exclusive or

p	q	$\mathit{p}\oplus \mathit{q}$
1	1	0
1	0	1
0	1	1
0	0	0

⇒	the if sign: A conditional statement read as “$if$ p, then q”

Table 3. Truth table of implication

p	q	$\mathit{p}\Rightarrow \mathit{q}$
1	1	1
1	0	0
0	1	1
0	0	1

Table 3. Truth table of implication

p	q	$\mathit{p}\Rightarrow \mathit{q}$
1	1	1
1	0	0
0	1	1
0	0	1

⇔	the if and only if sign: A biconditional statement read as “p $if$ $and$ $only$ $if$ q”

Table 4. Truth table of biconditional

p	q	$\mathit{p}\iff \mathit{q}$
1	1	0
1	0	1
0	1	1
0	0	0

Table 4. Truth table of biconditional

p	q	$\mathit{p}\iff \mathit{q}$
1	1	0
1	0	1
0	1	1
0	0	0

$\{x\in \mathbb{P}$|$x^2\}$	the given that sign: This set consists of $x^2$ values, $where$ x is a prime number
a|b	the divides sign: a $divides$ b without a remainder
∴$a\in \mathbb{K}$	the therefore sign: $Therefore$, a is an element of the set $\mathbb{K}$
a $\mathrm{R}$ b⇔a|b	the relation sign: a $related$ with b if and only if a divides b
$\forall a(a+1>0)$	the universal quantifier: $For$ $every$ a $number$, $a+1$ is greater than 2
$a_b$  $\in \mathbb{K}$	the sequence sign: A $sequence$ of numbers from the set $\mathbb{K}$, $a_1=x$, ..., $a_b=z$

2. Theorems and Proofs

Preliminary Theorem: For all distinct prime numbers p and for all distinct positive natural numbers n, then $p·n+p$ is a composite number.

$$p\in \mathbb{P}$$

$$n\in {\mathbb{N}}^{+}$$

$$(\forall p)(\forall n)(p·n+p)\in \mathbb{C}$$

Preliminary Proof: The expression p·$n+p$ can be written as p·$(n+1)$by distributive property1.

$$p·n+p=p·(n+1) \mathbf{by} \mathbf{distributive} \mathbf{property}$$

If n is any positive natural number and 1 is a positive natural number, then $n+1$ equals to be a positive natural number by the closure property of positive natural numbers under addition2. (n∈${\mathbb{N}}^{+}$∧ 1 ∈${\mathbb{N}}^{+})$⇒$(n+1$∈${\mathbb{N}}^{+})$ by closure property of positive natural numbers under addition

If n is any positive natural number, then $n+1$ equals to be either the composite number or the prime number, depending on Basis II.

 

      ($n+1$∈${\mathbb{N}}^{+})$⇒$\left(\right(n+1$∈$\mathbb{P})$⊕$(n+1$∈$\mathbb{C}\left)\right)$ by Basis II.

 

Let $q_x$ be a prime number and $c_x$ be a composite number. If $n+1$ is a prime number, then $p·(n+1)$ equals to be $p·$$q_x$. If $n+1$ is a composite number, then $p·(n+1)$ equals to be $p·$$c_x$.

$q_x$∈$\mathbb{P}$

$c_x$∈$\mathbb{C}$

 

    ($n+1$∈$\mathbb{P})$⇒$(p$·$(n+1)=p$·$q_x)$⊕$(n+1$∈$\mathbb{C})$⇒$(p$·$(n+1)=p$·$c_x)$

 

The result of the expression $p·$$q_x$ has divisors p and $q_x$ besides itself and 1. So the expression $p·$$q_x$ is a composite number according to Def. II. The result of the expression $p·$$c_x$ has divisors p and $c_x$ besides itself and 1. So the expression $p·$$c_x$ is a composite number according to Def. II. Therefore, for all distinct prime numbers p and for all distinct positive natural numbers n, then $p·(n+1)$, that is, $p·n+p$ equals a composite number.

a $\mathrm{R}$ p·$q_x$ ⇔ a|p·$q_x$

$(a=$ $q_x)$ ⇒ $\mathrm{R}$ $=\{1,$ p·$q_x$, p, $q_x\}$

$(a=$ $c_x)$ ⇒ $\mathrm{R}$ $=\{1,$ p·$c_x$, p, $c_x\}$

∴$(\forall p)(\forall n)(p·(n+1)=p·n+p)$∈$\mathbb{C}$by Def. II

Theorem I: Let p be a prime number and n be a positive natural number. Then the formula $p·n+p$ returns all composite numbers .

$\mathbb{C}=\{p$∈$\mathbb{P}$, n∈${\mathbb{N}}^{+}$|p·$n+p\}$

Proof I: As a result of the work on the proof of the preliminary theorem, the following can be said: with p prime, $n+1$ either prime or composite, this operation can be expressed as: “prime × (either prime or composite)”. So there are two possibilities:

Possibilites I) If $n+1$ is a composite number, then $p·(n+1)$ can be expressed as “prime × composite”. Adhering to Basis I, this expression can be written as “prime × (prime $\times \cdots \times$ prime)”.

c∈$\mathbb{C}$

$(n+1$∈$\mathbb{C})$⇒$(p·(n+1)=p·c)$

sec.3Basis I:

j, k∈${\mathbb{N}}^{+}$

$t_j$, $q_k$∈$\mathbb{P}$

$a=$ $t_1$·$t_2$$·\cdots ·$$t_j$

$b=$ $q_1$·$q_2$$·\cdots ·$$q_k$

$c=a·b$

∴$p·c=p·a·b=p$·$(t_1$·$t_2$$·\cdots ·$$t_j)$·$(q_1$·$q_2$$·\cdots ·$$q_k$)

Possibilities II) If $n+1$ is a prime number, then $p·(n+1)$ can be expressed as “prime× (prime)”.

x∈${\mathbb{N}}^{+}$

$p_x$∈$\mathbb{P}$

$(n+1$∈$\mathbb{P})$⇒$(p·(n+1)=p$·$p_x)$

As a result, there are two potential outcomes, “prime × (prime $\times \cdots \times$ prime)” and “prime × (prime)”. Since these two results are potentially the product of all prime numbers; so it can be said, adhering to Basis I, that this formula $p·n+p$ returns all composite numbers.

$(n+1$∈$\mathbb{C})$⇒$(p·(n+1)=p$·$(t_1$·$t_2$$·\cdots ·$$t_j)$·$(q_1$·$q_2$$·\cdots ·$$q_k\left)\right)$⊕$(n+1$∈$\mathbb{P})$

⇒$(p·(n+1)=p·$$p_x)$

sec.Basis I:

∴$\mathbb{C}=\{p$∈$\mathbb{P}$, n∈${\mathbb{N}}^{+}$|$p·n+p\}$

Since there are no even prime numbers except for two, the following theorem can be established to obtain only odd composite numbers:

Theorem II: Let p be a prime number and n be a positive natural number. Then the formula $2·n·p+p$ returns all odd composite numbers.

Proof II: The formula $p·n+p$ obtained in Theorem I has 4 possibilities arising from the fact that p can be even/odd and n can be even/odd:

Table 5. Table of all possibilities

Options	p	n	$\mathit{p}·(\mathit{n}+1)$
Option I	even	even	even· (even + 1)
Option II	even	odd	even· (odd + 1)
Option III	odd	even	odd· (even + 1)
Option IV	odd	odd	odd· (odd + 1)

Table 5. Table of all possibilities

Options	p	n	$\mathit{p}·(\mathit{n}+1)$
Option I	even	even	even· (even + 1)
Option II	even	odd	even· (odd + 1)
Option III	odd	even	odd· (even + 1)
Option IV	odd	odd	odd· (odd + 1)

Option I) Let k, t, z, and m be positive natural numbers. If p is an even number and n is an even number, then let p and n be equal to $2k$ and $2t$, respectively, according to the definition of even numbers mentioned in Preliminary Information II. If p is $2k$ and n is $2t$, then $p·(n+1)$ equals to be $2k·(2t+1)$. Let $2t+1$ be equal to z. If $2t+1$ is z, then $2k·(2t+1)$ equals to be $2kz$. Let $kz$ be equal to m. If k is m, then $2kz$ equals to be $2m$, which is an even number by the definition of even numbers. Therefore, if p is $2k$ and n is $2t$, then $p·(n+1)$ is an even number.

$k,t,z,m$∈${\mathbb{N}}^{+}$

$(p$∈$\mathbb{E})$⇒$(p=2k)$by def. of even numbers

$(n$∈$\mathbb{E})$⇒$(n=2t)$by def. of even numbers

$\left(\right(p=2k)$∧$(n=2t))$⇒$(p·(n+1)=2k·(2t+1\left)\right)$

$(2t+1=z)$⇒$(2k·(2t+1)=2kz)$

$(kz=m)$⇒$(2kz=2m$∈$\mathbb{E})$by def. of even numbers

∴$\left(\right(p=2k)$∧$(n=2t))$⇒$(p·(n+1)$∈$\mathbb{E})$

Option II) If p is an even number, then let p be equal to $2k$, and if n is an odd number, then let n be equal to $2t+1$ according to the definition of even numbers and odd numbers mentioned in Preliminary Information II. If p is $2k$ and n is $2t+1$, then $p·(n+1)$ equals to be $2k·(2t+1+1)$ equals to be $2k·(2t+2)$. Let $2t+2$ be equal to z. If $2t+2$ is z, then $2k·(2t+2)$ equals to be $2kz$. Let $kz$ be equal to m. If $kz$ is m, then $2kz$ equals to be $2m$, which is an even number by the definition of even numbers. Therefore, if p is $2k$ and n is $2t+1$, then $p·(n+1)$ is an even number.

$(p$∈$\mathbb{E})$⇒$(p=2k)$by def. of even numbers

$(n$∈$\mathbb{O})$⇒$(n=2t+1)$by def. of odd numbers

$\left(\right(p=2k)$∧$(n=2t+1))$⇒$(p·(n+1)=2k·(2t+1+1)=2k·(2t+2\left)\right)$

$(2t+2=z)$⇒$(2k·(2t+2)=2kz)$

$(kz=m)$⇒$(2kz=2m$∈$\mathbb{E})$by def. of even numbers

∴$\left(\right(p=2k)$∧$(n=2t+1))$⇒$(p·(n+1)$∈$\mathbb{E})$

Option III) If p is an odd number, then let p be equal to $2k+1$, and if n is an even number, then let n be equal to $2t$ according to the definition of odd numbers and even numbers mentioned in Preliminary Information II. If p is $2k+1$ and n is $2t$, then $p·(n+1)$ equals to be $(2k+1)·(2t+1)$ equals to be $(2k+1)·(2t+1)$ equals to be $4kt+2k+2t+1$ equals to be $2·\left(2kt\right)+2·(k+t)+1$. Let $2kt$ and $k+t$ be equal to z and m, respectively. If $2kt$ is z and $k+t$ is m, then $2·\left(2kt\right)+2·(k+t)+1$ equals to be $2z+2m+1$ equals to be $2·(z+m)+1$. Let $z+m$ be equal to a. If $z+m$ is a, then $2·(z+m)+1$ equals to be $2a+1$, which is an odd number by the definition of odd numbers. Therefore, if p is $2k+1$ and n is $2t$, then $p·(n+1)$ is an odd number.

a∈${\mathbb{N}}^{+}$

$(p$∈$\mathbb{O})$⇒$(p=2k+1)$by def. of odd numbers

$(n$∈$\mathbb{E})$⇒$(n=2t)$by def. of even numbers

$\left(\right(p=2k+1)$∧$(n=2t))$

⇒$(p·(n+1)=(2k+1)·(2t+1)=4kt+2k+2t+1=2·(2kt)+2·(k+t)+1)$

$\left(\right(2kt=z)$∧$(k+t=m))$⇒$(2·(2kt)+2·(k+t)+1=2z+2m+1=2·(z+m)+1)$

$(z+m=a)$⇒$(2·(z+m)=2a+1$∈$\mathbb{O})$by def. of odd numbers

∴$\left(\right(p=2k+1)$∧$(n=2t))$⇒$(p·(n+1)$∈$\mathbb{O})$

Option IV) Let a be a positive natural number. If p is an odd number and n is an odd number, then let p and n be equal to $2k+1$ and $2t+1$, respectively, according to the definition of odd numbers mentioned in Preliminary Information II. If p is $2k+1$ and n is $2t+1$, then $p·(n+1)$ equals to be $(2k+1)·(2t+1+1)$ equals to be $(2k+1)·(2t+2)$ equals to be $4kt+4k+2t+2$ equals to be $2·\left(2kt\right)+2·\left(2k\right)+2t+2$. Let $2kt$ and $2k$ be equal to z and m, respectively. If $2kt$ is z and $2k$ is m, then $2·\left(2kt\right)+2·\left(2k\right)+2t+2$ equals to be $2z+2m+2t+2$ equals to be $2·(z+m+t+1)$. Let $z+m+t+1$ be equal to a. If $z+m+t+1$ is a, then $2·(z+m+t+1)$ equals to be $2a$, which is an even number by the definition of even numbers. Therefore, if p is $2k+1$ and n is $2t+1$, then $p·(n+1)$ is an even number.

$(p$∈$\mathbb{O})$⇒$(p=2k+1)$by def. of odd numbers

$(n$∈$\mathbb{O})$⇒$(n=2t+1)$by def. of odd numbers

$\left(\right(p=2k+1)$∧$(n=2t+1))$

⇒$(p·(n+1)=(2k+1)·(2t+1+1)=(2k+1)·(2t+2)=4kt+4k+2t+2=(2kt)+2·(2k)+2t+2)$

$\left(\right(2kt=z)$∧$(2k=m))$⇒$(2·(2kt)+2·(2k)+2t+2=2z+2m+2t=2·(z+m+t+1\left)\right)$

$(z+m+t+1=a)$⇒$(2·(z+m+t+1)=2a$∈$\mathbb{E})$by def. of even numbers

∴$\left(\right(p=2k+1)$∧$(n=2t+1))$⇒$(p·(n+1)$∈$\mathbb{E})$

As a result, a table like this can be created:

Table 6. Table of results of all possibilities

Options	p	n	$\mathit{p}·(\mathit{n}+1)$
Option I	even	even	even
Option II	even	odd	even
Option III	odd	even	odd
Option IV	odd	odd	even

Table 6. Table of results of all possibilities

Options	p	n	$\mathit{p}·(\mathit{n}+1)$
Option I	even	even	even
Option II	even	odd	even
Option III	odd	even	odd
Option IV	odd	odd	even

The third option is the only way to return an odd number from the formula that gives all composite numbers. In this case, a formula that satisfies the requirements of the third option returns all odd composite numbers. Because the third option is derived from the formula that gives all composite numbers, and a result is always an odd number. The requirements of the third option are that the prime number p is an odd prime number and the positive natural number n is a positive even natural number in the formula $p·(n+1)$. In this context, the number $n+1$ is also odd. Let p be an odd number to generate the formula that returns only odd composite numbers. Provided that p is constant, one composite number must be obtained for each natural number. That’s why we need to manipulate the $n+1$ value in the formula $p·(n+1)$ so that it is always an odd number. So if we multiply n by 2 for satisfying the requirements of the third option, we get $p·(2n+1)$, where $2n+1$ is an odd number according to the definition of odd numbers in Preliminary Information II. For all distinct prime numbers p, as proved above, the formula $p·(n+1)$ returns all odd composite numbers as long as n is any positive even natural number; if we want the formula to return all odd composite numbers for each positive natural number, that is, to be more optimized, the formula $p·(2n+1)$ derived from the formula $p·(n+1)$ gives all odd composite numbers as long as n is any positive natural numbers.

3. Results

For the potential of the relationship between composite numbers and prime numbers to provide solutions to mathematical problems related to prime numbers, we have presented three theorems, one of which is a preliminary theorem, and one proof for each theorem, a total of three proofs.

As a result, we proved that formula $n·p+p$ returns all composite numbers, and the formula $2·n·p+p$ returns all odd composite numbers.

In addition, we explained the similarities and differences between the formula we developed with the previously known formulas.