TITLE: Why is $ab$ likely to have more divisors than $(a-b)(a+b)$?
QUESTION [19 upvotes]: Consider the two numbers $ab$ and $(a-b)(a+b), \gcd(a,b) = 1, 1 \le b < a$. On an average, which of these two numbers has more distinct prime factors? All the prime factors of $a$ and $b$ divide $ab$ and similarly all the prime factors of $a-b$ and $a+b$ divide $(a-b)(a+b)$. So one number does not seem to have an obvious advantage over the other. However if we look at the data than we see that $ab$ dominates.
Let $f(x)$ be the average number of distinct prime factors in all such $ab, a \le x$ and
$g(x)$ be the average number of distinct prime factors in all such $(a-b)(a+b), a \le x$.
Update: Experimental data for the first $6.1 \times 10^{9}$ pairs of $(a,b)$ shows that $f(x) - g(x) \sim 0.30318$. Instead of distinct prime factors, if we count the number of divisors than $f(x) - g(x) \sim 0.848$.
Question: Why is $ab$ likely to have more distinct prime factors or divisors than $(a-b)(a+b)$ and what is the limiting value of $f(x) - g(x)$?

REPLY [2 votes]: In case A, looking at the set $ \{(1,0),(0,1),(1,1)\} $, the probability that 2|ab is 2/3
In case B, looking at the same set, the probability that $ 2|(a-b)(a+b) $ is $ 1/3
 $
For all other odd primes, the probability that $p|ab$ is the same as $p|(a-b)(a+b)$, and is: $2(p-1)/(p^2-1) = 2/(p+1)$
However, for large primes, $a<p<2a$, it is possible that $p|(a+b)$ but never $p|ab$.
Hence, the result that comes out empirically is somewhat less than 1/3.
At 1/3 of the average, some of the a+b=p mentioned must be subtracted.