Port563
Posts:
676
Registered:
12/15/13


Tough problem with primes?
Posted:
Dec 15, 2013 2:57 AM


THEOREM ************ For any six consecutive natural numbers, there exists a prime such that it divides exactly one of those naturals. ************
I haven't been able to prove this.
This is what I managed so far 
As we have 6 consecutive naturals, it is obvious that both 2 and 3 (the first primes) are always going to divide more than one of them, and so cannot ever be candidate primes to do the job.
The smallest candidate prime that can possibly do the job for six consecutive naturals is 5.
If we examine the first few sets of six consecutive natural numbers, e.g. {1,2,3,4,5,6}  5 does it {2,3,4,5,6,7}  5 does it {3,4,5,6,7,8}  5 does it (so does 7) {4,5,6,7,8,9}  5 does it (so does 7) {5,6,7,8,9,10}  7 does it {6,7,8,9,10,11}  5 does it (so do 7 & 11) a pattern becomes clear.
It is trivial to show that for all sets of 6 consecutive naturals {k,k+1,k+2,k+3,k+4,k+5}, where k is a natural that is NOT divisible by 5, then 5 will divide exactly one of the elements of the set (and will "do the job").
Therefore, all that remains is to prove the result for all {5m,5m+1,5m+2,5m+3,5m+4,5m+5}, where m is a natural.
It is also trivial to prove that 7 will do the job for such a set if and only if 5m1 is not divisible by 7.
So the residue is the infinity of sets  {15,16,17,18,19,20} {50,51,52,53,54,55} {85,86,87,88,89,90} (120,121,122,123,124,125} etc. Neither 5 nor 7 can divide exactly one element of them (for each set, 5 will divide two elements and 7 none).
Thus, in order to prove the theorem, all that remains is to prove that for every such set, i.e.
{35n20,35n19,35n18,35n17,35n16,35n15}
where n is natural, a prime > 7 exists that divides one of the elements. (As it is trivial to prove that if such a prime divides one element of such a set, it cannot possibly divide any other element, I replaced "exactly one" with "one")
If n=1, 17 does it (so does 19) If n=2, 11 does it (so do 13, 17 & 53) If n=3, 11 does it (so do 17, 29, 43 & 89) If n=4, 11 does it (so do 31, 41 & 61)
Indeed, after getting stuck at this point, just in case this was a practical joke (i.e., the theorem isn't true, and so proving it would be rather difficult) I computerchecked that a prime divisor > 7 exists for all n up to 10^6. In doing so, I never had to use a prime above 7919 (the 1,000th prime).
I can't see how to apply either induction or reductio ad absurdum here, to prove it true for all n.
Fermat's (little) theorem, which states that if p is prime, for any integer n, (n^p  n) is divisible by p, also doesn't seem to help.
I tried many other promising things, but they all led up blind alleys.
I'm obviously missing something  suggestions, please!

