Are there arbitrarily long prime gaps in which each number has at least three distinct prime factors?

Solution 1:

Yes, there are arbitrarily long sequences of integers each of which has at least $3$ prime factors, and the same is true for any $k$ in lieu of $3$.

Given arbitrary positive integers $k$ and $m$, choose integers $n_1,n_2,\ldots,n_m$ with the following properties:

1. Each $n_i$ has at least $k$ distinct prime factors;
2. The $n_i$ are pairwise relatively prime.

It is obvious that such a sequence of integers exists (just pick a batch of distinct primes for each $i$, and multiply them).

Now, using the Chinese Remainder Theorem we can find an integer $N$ such that for each $i$, $N \equiv -i \pmod{n_i}$, or in other words $N+i \equiv 0 \pmod{n_i}$. Therefore, the numbers $N+1, N+2, \ldots, N+m$ all have at least $k$ distinct prime factors.

EDIT: the question seems to further demand that the long stretch of “very composite” integers be bookended by actual primes. This proof does not guarantee that, and in fact I suspect this is very hard to guarantee.