Equivalence of Definitions of Prime Number

Definition 1 iff Definition 2
This is proved in Prime Number has 4 Integral Divisors:

Definition 1 iff Definition 3
This is proved in Divisor Counting Function of Prime Number:

Definition 1 iff Definition 4
From these two results:
 * $1$ Divides all Integers
 * Integer Divides Itself

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.

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 $p$.

Definition 2 iff Definition 6
This is proved in Prime iff Equal to Product.

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 $p$.