I've been recently reading Yuan Wang's paper on an approximation to Goldbach's problem, in which he showed that
Proposition 1: For all large even integer $x$, there exists $1<n<x-1$ such that $n(x-n)$ has no prime divisors $\le x^{1/8}$, at most three prime divisors in $x^{1/8}<p\le x^{1/2}$ such that $p^2\nmid n(x-n)$.
From this, he concluded without elaborations that
Proposition 2: For all large even integer $x$, there exists $1<n<x-1$ such that $n(x-n)$ has at most five prime factors.
However, the best I can do from proposition 1 is that $n(x-n)$ has at most six prime factors:
Let $q$ be the product of all $m$ prime divisors of $N=n(x-n)$ that are $\le x^{1/2}$. Then
$$ x^{m/8}<q\le x^{m/2} $$
Moreover
$$ {N\over x^{m/2}}\le\frac Nq<{N\over x^{m/8}}<x^{2-\frac m8} $$
By definition, we know that all prime divisors of $N/q$ are $>x^{1/2}>(N/q)^{1\over2(2-m/8)}:=(N/q)^{1/s}$. Now, all we need is to find out the number of prime divisors of $N$ based on different values of $m$. For all $1\le m\le3$, wee see that $3<s<4$, so $N/q$ has at most three prime divisors. This indicates that $N$ should only have at most six prime divisors.
This made me confused about roles of $m\le3$ in his paper, so I wonder whether anyone could help me understand this better.
Since $n,x-n<x$, there can be at most two prime factors of $N=n(x-n)$ that are $\geq x^{1/2}$ by the pigeonhole principle, as each must divide either $n$ or $x-n$. Assuming $p^2\not\mid N$ for $p\mid N$, $p\leq x^{1/2}$ as you said (I failed to find this in the paper but I did not search thoroughly), we find that there can be at most $5$ prime factors of $N$.