Consider six consecutive positive integers. Show that there is a prime number that divides exactly one of them.

I am more inclined than ever to believe Steve Herman's conjecture:

"For n>1, there is a prime number that divides exactly one of any n consecutive positive integers."

Here's my thinking, with some hand-waving:

Define JLo's condition:  n consecutive numbers satisfy JLo's condition if some prime number <=n does not divide at least two of the numbers or if at least one of the numbers is not divisible by a prime <=n.

Observation: If a sequence satisfies JLo's condition, then it has at least one prime number that divides exactly one of them

Observation:  For n <= 19, we can prove that all sequences satisfy JLo's condition.

Handwaving: For n >= 20, any sequence that does not satisfy JLo's condition must have a sufficiently large starting number that at least one of it's terms (and in fact, most of its' terms) have a prime factor > n.  That prime factor perforce divides only one term in the sequence.  For instance, for n = 25, the first sequence that does not satisfy JLo's condition starts with something like 185,065,429.  Every one of the 13 odd terms in this sequence (and in any 25-length sequence which do not satisfy JLo's condition) are provably divisible by exactly one prime < 25.  Unless all of 13 them are exact powers of one of the 9 primes less than 25, then one of them has a prime factor > 25.  It is impossible for all of them to be exact powers of one of the 8 primes less than 25, so there is a prime number that divides exactly one of any 25 consecutive positive integers.

So, I know that I can prove Steve Herman's conjecture for n= 20, 21, 22, 23, 24, 25.  It might be possible to take an approach like this and formalize and actually prove this for all n, but I'm not going to be doing it.  And it is also possible that the conjecture is true but unprovable.

Posted by Steve Herman on 2006-09-09 21:03:21