I recently asked a question on MathOverflow that got me thinking about Legendre's Conjecture.
Consider a range of consecutive integers defined by $R(x+1,x+n) = x+1, x+2, x+3, \dots, x+n$ with $C(x+1,x+n)$ = the count of distinct least prime factors for $R(x+1,x+n)$
For a given $n$, let $p_k$ be the $k$th prime that is the greatest prime less than or equal to $n$ so that $p_k \le n < p_{k+1}$
Would proving that for $x \ge 1,n \ge 4, C(x+1, x+2n) \ge k+2$ be enough to prove Legendre's Conjecture?
Here's my thinking:
(1) Assume that Legendre's Conjecture is false. Let $n$ be the first time that the conjecture fails. So, for any $i$ where $n^2 < i < n^2+2n+1$, $i$ is not prime.
(2) If $C(x+1, x+2n) \ge k+2$ for all $x \ge 1, n \ge 4$, then $C(n^2+1, n^2+2n) \ge k+2$
(3) But then, there must be a prime greater than $p_{k+1}$ that is a least prime factor for some $i$ where $n^2 < i < n^2 + 2n + 1$. But this is impossible since the first time a non-prime can have a least prime factor greater or equal to $p_{k+2}$ is $(p_{k+2})^2 > (n+1)^2$
(4) Since, we have a contradiction, we can reject our assumption in step(1).
Is this reasoning correct?
Your reasoning is correct, but here's what I think is a bit simpler (or, at least, different) way to express it. Your conjecture states
If your conjecture is correct, but Legendre's conjecture is false, then choosing an $n$ which fails and $x = n^2$, we get your point $(2)$, i.e.,
Since there's no prime $p$ where $n^2 + 1 \le p \le n^2 + 2n$, then each integer in that range is composite, so each value must be at least the square of its least prime factor. Note that
$$n \lt p_{k+1} \;\to\; n + 1 \le p_{k+1} \;\to\; (n + 1)^2 \le p^2_{k+1} \;\to\; n^2 + 2n \lt p^2_{k+1}$$
so actually $C(n^2+1, n^2+2n) \le k$. However, this contradicts your point $(2)$ (actually, just proving $C(n^2+1,n^2+2n) \ge k+1$ would be sufficient). Thus, if your conjecture is true, then the assumption that Legendre's conjecture is false must be incorrect (i.e., Legendre's conjecture would be true).