# Equivalence of Definitions of Prime Number

## Theorem

The following definitions of the concept of **Prime Number** are equivalent:

### Definition 1

A **prime number** $p$ is a positive integer that has exactly two divisors which are themselves positive integers.

### Definition 2

Let $p$ be a positive integer.

Then $p$ is a prime number if and only if $p$ has exactly four integral divisors: $\pm 1$ and $\pm p$.

### Definition 3

Let $p$ be a positive integer.

Then $p$ is a prime number if and only if:

- $\map {\sigma_0} p = 2$

where $\map {\sigma_0} p$ denotes the divisor count function of $p$.

### Definition 4

A **prime number** $p$ is an integer greater than $1$ that has no positive integer divisors other than $1$ and $p$.

### Definition 5

A **prime number** $p$ is an integer greater than $1$ that has no (positive) divisors less than itself other than $1$.

### Definition 6

Let $p \in \N$ be an integer such that $p \ne 0$ and $p \ne \pm 1$.

Then $p$ is a **prime number** if and only if

- $\forall a, b \in \Z: p \divides a b \implies p \divides a$ or $p \divides b$

where $\divides$ means **is a divisor of**.

### Definition 7

A **prime number** $p$ is an integer greater than $1$ which cannot be written in the form:

- $p = a b$

where $a$ and $b$ are both positive integers less than $p$.

## Proof

### Definition 1 iff Definition 2

This is proved in Prime Number has 4 Integral Divisors:

### Necessary Condition

Let $p$ be a prime number from the definition that $p$ has exactly $2$ divisors which are positive integers.

From One Divides all Integers and Integer Divides Itself those positive integers are $1$ and $p$.

Also, we have $-1 \divides p$ and $-p \divides p$ from One Divides all Integers and Integer Divides its Negative.

Aiming for a contradiction, suppose:

- $\exists x < 0: x \divides p$

where $x \ne -1$ and $x \ne -p$.

Then:

- $\size x \divides x \divides p$

and so $\size x$ is therefore a positive integer other than $1$ and $p$ that divides $p$.

This is a contradiction of the condition for $p$ to be prime.

So $-1$ and $-p$ are the only negative integers that divide $p$.

It follows that $p$ has exactly those four divisors.

$\Box$

### Sufficient Condition

Suppose $p$ has the divisors $1, -1, p, -p$.

It follows that $1$ and $p$ are the only positive integers that divide $p$.

Thus $p$ has exactly two divisors which are positive integers.

$\blacksquare$

### Definition 1 iff Definition 3

This is proved in Divisor Count Function of Prime Number:

### Necessary Condition

Let $p$ be a prime number.

Then, by definition, the only positive divisors of $p$ are $1$ and $p$.

Hence by definition of the divisor count function:

- $\map {\sigma_0} p = 2$

$\Box$

### Sufficient Condition

Suppose $\map {\sigma_0} p = 2$.

Then by One Divides all Integers we have:

- $1 \divides p$

Also, by Integer Divides Itself we have:

- $p \divides p$

So if $p > 1$ it follows that $\map {\sigma_0} p \ge 2$.

Now for $\map {\sigma_0} p = 2$ it must follow that the only divisors of $p$ are $1$ and $p$.

That is, that $p$ is a prime number.

$\blacksquare$

### Definition 1 iff Definition 4

From these two results:

it follows that if $p$ has exactly two positive integer divisors then those are $1$ and $p$.

By the same coin, if the only positive integer divisors of $p$ are $1$ and $p$, then $p$ has exactly two positive integer divisors.

$\blacksquare$

### Definition 4 iff Definition 5

Let the only two positive integer divisors of $p$ be $1$ and $p$.

Then the only divisor of $p$ strictly less than $p$ is $1$.

Conversely, let the only divisor of $p$ strictly less than $p$ be $1$

From Integer Divides Itself we also have that $p$ is a divisor of $p$.

From Absolute Value of Integer is not less than Divisors it follows that any positive integer greater than $p$ is not a divisor of $p$.

Thus the only positive integer divisors of $p$ are $1$ and $p$.

$\blacksquare$

### Definition 2 iff Definition 6

This is proved in Prime iff Equal to Product.

$\blacksquare$

### Definition 4 iff Definition 7

Let the only two positive integer divisors of $p$ be $1$ and $p$.

Then the only positive integer less than $p$ which divisors of $p$ is $1$

So if $a b = p$ then either $a$ or $b$ is $p$ and so it is not the case that both $a$ and $b$ are less than $p$.

Now suppose that there do not exist $a$ and $b$ less than $p$ such that $a b = p$.

Suppose $a \divides p$ such that $a < p$.

That means that $\exists b \in \Z_{>0}: a b = p$.

That means $b \divides p$.

But $b \not < p$ by hypothesis.

It follows that $b = p$ and so $a = 1$

Hence $1$ and $p$ are the only positive integer divisors of $p$.

$\blacksquare$

## Sources

- 1992: George F. Simmons:
*Calculus Gems*... (previous) ... (next): Chapter $\text {B}.2$: More about Numbers: Irrationals, Perfect Numbers and Mersenne Primes: Footnote $1$