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Let P(m,n) $P(m,n)$ mean that there is a number, M, $M$, such that starting with M $M$ there are m $m$ consecutive numbers each having exactly n $n$ distinct prime factors. Is it obvious that P(m,n) $P(m,n)$ is true for all m $m$ and n? $n$? My gut says "obviously" and P(4,4) $P(4,4)$ and P(5,5) $P(5,5)$ are definitely true (for 134043 and 129963314 respectively). It seems like some sort of pigeonhole proof based on the number of factors available might work, but upon reflection, I'm not so sure. Maybe I'm missing something obvious.