Primality of Second Smallest Divisor

Let $d$ be the 'second' divisor of $n$ , which means, there is no integer $m$ with $1<m<d$ that $m | n$ .

Now, $d$ must be a prime; otherwise we can find such an $m$ . If $d$ is not a prime, then there is at least one integer $x$ with $1<x<d$ that $x | d$ . As $x | d$ and $d | n$ , so we have, $x | n$ with $1<x<d$ ; which makes $x$ a fine $m$ .

So, the conjecture is $\boxed{\text{True for Every } n}$ .

Given { $1, b, c, d, e\ldots N$ } are the factors of $N$ in increasing order from $1$ to $N$ .

Assume $b$ is not a prime, then it can be factored into $f_1 \cdot f_2 \cdot f_3 \cdot \ldots$ which all must be smaller than $b$ . So taking the multiple of $b$ that gives $N$ , call it $q$ (meaning $\frac{N}{b}=q$ ), then $f_1\cdot\frac{b\cdot q}{f_1}=f_1\cdot \frac{f_1 \cdot f_2 \cdot f_3 \cdot \ldots \cdot q}{f_1}=N$ so $f_1$ is a factor of $N$ smaller than $b$ which contradicts our claim that the non-prime number $b$ is the second smallest factor so it must be prime .

Let the divisors of $n$ be written in ascending order. Such that $n: 1, {\color{#D61F06}x}, ...$ where ${\color{#D61F06}x}$ is not a prime number.

It follows that, due to the Fundamental Theorem of Arithmetic, $x$ can be written as a unique product of primes which are necessarily smaller than $x$ .

We can rewrite the ordered list of the divisors of $n$ which includes the prime factors of $x$ and therefore if the divisors of any integer are listed in ascending order, the second divisor in the list will be prime.