[Edited following Gerhard's answer. I had forgotten to state a crucial assumption; my apologies for the confusion. I have reworded slightly to try to make things clearer.]
Let $p$ be the $k$-th odd prime. Let $n$ be an even integer. Let $p_{1}$, $\ldots$, $p_{k+1}$ be consecutive odd primes. Suppose that $p$ divides $n - p_{i}$ for all $1 \leq i \leq k+1$. For instance, take $p=3$, take $n=100$, take $p_{1}=31$, and take $p_{2}=37$. Is there then at least one $i$ such that $n - p_{i}$ has a prime factor strictly greater than $p$?
Anything anybody can say (proof or reference) is welcome.
It is easy to prove that this holds when $k=1$. Indeed, using the fact that $n$ is even, so that both $n - p_{1}$ and $n - p_{2}$ are odd, the claim is immediate unless both $n-p_{1}$ and $n-p_{2}$ are powers of 3. In that case, the difference between $p_{1}$ and $p_{2}$ is $2p_{1}$, and this is impossible by Bertrand's postulate.
There are also plenty of examples when $k=2$. The one with lowest $n$ is $n=1662$, and the primes $1627$, $1637$, $1657$. Experimental evidence suggests that the question can be answered in the affirmative in this case too.
I would guess that there are examples for all $k$, but that very large numbers will be required; I do not actually have an example for any $k > 2$. Anything that can be said about the existence of examples is welcome too.
