# Prime factors in a range of composite integers

The specific question I am interested in is the following. Let $p_{1}$, $\ldots$, $p_{k}$ be the first $k$ odd primes in succession, and let $n$ be some even integer strictly greater than $p_{k}$. Suppose that $n - p_{i}$ is not prime for any $1 \leq i \leq k$. Experimental evidence suggests that there must then be at least one $i$ with $1 \leq i \leq k$ such that $n - p_{i}$ has a prime factor which is greater than or equal to $p_{k}$. Is there a result in the literature which implies this, or can someone come up with a nice proof?

The consideration of the first $k$ primes here might conceivably be a distraction. What are related known facts about distinct prime factors that can occur in a certain range of composite integers? Grimm's conjecture implies that there are at least $k$ distinct prime factors in a list of consecutive integers of length $k$, but the facts I am looking for seem quite a bit weaker (fortunately, since Grimm's conjecture is expected to be very hard).

Additional comments (01/01/18): to give a little motivation, I will explain that if the above is true for a given $n$ and any $k$, it appears to imply the Goldbach conjecture for $n$. Indeed, suppose that for a fixed $n$ and a given $k$ as above, we have that $n - p_{i}$ is not prime for any $1 \leq i \leq k$. Suppose that we can answer the question I asked in the affirmative, so that there must then be a prime $p'$ greater than or equal to $p_{k}$ which is a prime factor of $n - p_{i}$ for some $i$. Then we can show that the following holds.

Claim. There is a prime which is strictly greater than $p_{k}$ and strictly less than $n$.

Proof of claim. If $p'$ itself is strictly greater than $p_{k}$, then we are done: since $p'$ is a prime factor of $n - p_{i}$, it is certainly strictly less than $n$. If $p'$ is equal to $p_{k}$, then Bertrand's postulate implies that there is a prime $p''$ such that $p' = p_{k} < p'' < 2p_{k}$. Since $n - p_{i}$ is not prime, it has at least one prime factor in addition to (and possibly the same as) $p_{k}$. We thus have that $n - p_{i} \geq 3p_{k}$, so that $p'' < n - p_{i} < n$.

The Goldbach conjecture follows immediately from the claim, by induction. Indeed, since there can be only finitely many primes less than $n$, there must be some integer $k$ to which we cannot apply the claim. For the first such $k$, this means that $n - p_{k}$ is prime, as required.

One might conclude from this that a proof of the question I asked must be very hard. But it doesn't seem that way to me, it seems that quite coarse estimates should be able to do it. I have some ideas, but these seem to be leading in a direction that is a bit more involved than it feels to me should be necessary. I am not an expert on sieve theory or similar things, and I am hoping that somebody who is might see a simple proof or know of a fact from which it follows.

• How about a nice (almost) counterexample? N=30 and k=2. Also, Langevin has a proof for large n. Cf. 262400 . Gerhard "Nice Way To Finish 2017" Paseman, 2017.12.31. – Gerhard Paseman Dec 31 '17 at 16:18
• Why "almost" counterexample? – Aaron Meyerowitz Dec 31 '17 at 16:23
• Because he asks for greater or equal. 25 has a prime factor of 5 which is at least $p_2$. If the poster asked for strictly greater then it would be a counterexample. Gerhard "Just Counting Odd Primes Here" Paseman, 2017.12.31. – Gerhard Paseman Dec 31 '17 at 16:25
• For fixed $k$ there are only finitely many $n$ for which $x=n-3,y=n-5$ are both divisible only by the first $k$ odd primes, since $x/2-y/2=1$ is a solution to the $S$-unit equation, with $S$ being the first $k+1$ primes. Assuming the abc conjecture, you can get a decent upper bound on $n$. – Felipe Voloch Dec 31 '17 at 18:51
• Really? More decent than the lcm of the odd numbers up to (p_k)/2? Gerhard "Prefers One Two Three Theorem" Paseman, 2017.12.31. – Gerhard Paseman Dec 31 '17 at 19:29