Let lpf($x)$ be the least prime factor of $x$.
It seems to me that if:
- $1 \le a \le n$
- $n \ge 128$
- $an < x \le (an+n)$
- lpf($x$) $\ge \frac{n}{4}$
Then, for all $y$ where:
- $an < y \le (an+n)$
- $ y \ne x$
It follows that
- lpf($y$) $< \frac{n}{4}$
Here's my reasoning:
Claim 1: if $p \ge \frac{n}{4}$, then $p^3 > n^2+n$
(1) $\frac{n}{64} \ge 2$
(2) $\frac{n^2}{64} \ge n + n > n + 1$
(3) $p^3 \ge \frac{n^3}{64} > n^2 + n$
Claim 2: if $p,q$ are primes where $p > q$ and $\frac{n^2}{16} < pq < n^2$, then $p^2 - q^2 > n$
(1) $(i+2)^2 - i^2 = 4i+4$
(2) $p^2 - q^2 \ge 4\left(\frac{n}{4}\right)+4 > n$
Claim 3: if $an < y \le (an+n)$ and $y \ne x$, then lpf($y$) $< \frac{n}{4}$
(1) abs($x-y) < $ abs($an+n - an) = n$
(2) Assume that $n >$ lpf($y$) $\ge \frac{n}{4}$
(3) Let $p = $lpf($y$)
(4) $y = p^2$
$p < y$ since $y > n > p$ and from Claim 1, $p^3 > y$
(5) By the same argument if $q = $lpf($x$), then $q^2 =x$
(6) But abs($x - y) =$ abs($p^2 - q^2) > n$ so the conclusion follows.
Is my reasoning correct? Is there any mistake in my thinking?
Edit 1: I realized an oversight.
Let $q,p$ be primes $p>q$. There is the possibility that $p=q+2$ which means that $pq - q^2 = 2q < n$.
So, the argument only applies to squares of distict primes but not to least prime factors.