TITLE: Is the smallest non-1 divisor of a number always prime?
QUESTION [18 upvotes]: I am working on some things related to arithmetic derivatives, and I was writing a program to calculate the arithmetic derivative. In working on my program I came across an assumption that I have made, but I could not prove that it was true and it was quite frustrating to me. 
My hypothesis is for any $n \in \mathbb{N}$, the smallest number, $a \in \mathbb{N}/\{1\}$ such that $a \vert n$ should be prime. My beginning of a proof for this was this
Using the fundamental theorem of arithmetic, for all $n \in \mathbb{N}/\{1\}$, that either $n$ is prime, or that $n$ has a unique prime factorization. If $n$ is prime, the proof is trivial, as the prime factorization for all $p \in \mathbb{P}$ is just $p$; however, for the case of $n$ being composite, the proof became quite difficult for me. 
For the case of a number $c$ such that $c \in \mathbb{N}/\{1\}$ and $c \notin \mathbb{P}$, we can say that there exists at least two numbers, $a$ and $b$ such that fulfill the following four statement

$1 < a \le b < c$
$a \vert c$
$b \vert c$
$ab = c$

My thought process thereby went into dividing $c$ into odds and evens. We can say that this theorem is true for all even numbers, $e$, rather trivially as well because for all $e \in \{2k:\mathbb{N}/\{1\}\}$ that $2\vert e$ and $2$ is also the smallest integer that could possibly fulfill the above criteria thereby making $a = 2$ and fulfilling the theorem for all even numbers.
The part that I cannot figure out however is odd numbers. My intuition is telling me this must be true, but I cannot figure out a proof of this theorem for odds and I was wondering if 

Is this theorem actually true?
If so, how can this be proven for odds?

REPLY [2 votes]: If you just want to prove that the smallest non 1 divisor of a number is always prime, all you have to do is proceed by contradiction. 
Let N be a composite number ( even or odd doesn't matter ). Let's assume that N's smallest non 1 divisor is not a prime, let's call this composite divisor c. 
If c is composite then using the fundamental theorem of arithmetic, c can be expressed as the factorisation of prime numbers all < to c. Let p be one of c's factors. 
Then since c | N and p | c then p | N , but p < c ( since p is a factor of c ), which contradicts the assumption that the smallest divisor of N is c. 
I don't know if this is what you wanted and if I have been clear enough.