Let $P$ be the probability that two integers where $m>1$is a fixed positive integer and $n$ is a randomly chosen such than $ m> n \geq 0 $? What is the probability $m^2 -n^2$ and be factored into $2$ factors.
I think the probability is at least $P(m^2 -n^2 = p_i p_j) \geq \frac{1}{m}$, where $p_i$ and $p_j$ are arbitrary primes.
On
According to OEIS entry A001358, the $n$th semiprime is asymptotically equal to $n \log(n) / \log ( \log (n))$.
This means the "probability" that a number $n$ is semiprime is asymptotically equal to $\log (\log (n))/ \log(n)$.
Therefore the probability that $m^2-n^2$ is semiprime for any natual $m, n$ is $\log (\log (m^2-n^2))/ \log(m^2-n^2)$.
I'm interpreting the question as: "For a fixed $m$, how many $0\le n<m$ have the property that $n-m$ and $n+m$ are both prime?" (Note that this is a little bit wrong for $n=m-1$, but never mind.)
This is exactly the same as asking how many representations $2m$ has as the sum of two primes. In other words, this is the Goldbach conjecture in disguise.
So we would love to prove (but currently can't) that this number of representations is positive for all $m\ge2$.
There is also a conjectured asymptotic formula, as a function of $m$, for the number of such representations.