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Topic: Tough problem with primes?
Replies: 52   Last Post: Jan 23, 2015 8:03 PM

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 Port563 Posts: 918 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 5m-1 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.

{35n-20,35n-19,35n-18,35n-17,35n-16,35n-15}

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 computer-checked 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!

Date Subject Author
12/15/13 Port563
12/15/13 quasi
12/15/13 Port563
12/15/13 quasi
12/15/13 Port563
12/15/13 quasi
12/16/13 Helmut Richter
12/16/13 Port563
12/16/13 Helmut Richter
12/16/13 Port563
12/16/13 Port563
12/20/13 m.michael.musatov@gmail.com
12/17/13 quasi
12/17/13 Wally W.
12/17/13 quasi
12/17/13 Helmut Richter
12/17/13 quasi
12/17/13 quasi
12/28/13 Wizard
12/28/13 Helmut Richter
12/28/13 Wizard-Of-Oz
12/28/13 Wizard-Of-Oz
12/28/13 Port563
12/28/13 Wizard-Of-Oz
12/29/13 Port563
12/17/13 gnasher729
12/17/13 quasi
12/17/13 Helmut Richter
12/17/13 Helmut Richter
12/19/13 Martin Shobe
12/29/13 Wizard-Of-Oz
1/23/15 I Am Me
12/17/13 Port563
12/17/13 quasi
12/18/13 Brian Q. Hutchings
12/18/13 quasi
12/19/13 quasi
12/19/13 quasi
12/19/13 Port563
12/19/13 Martin Shobe
12/19/13 Helmut Richter
12/19/13 Martin Shobe
12/19/13 quasi
12/19/13 quasi
12/22/13 Helmut Richter
12/22/13 Martin Shobe
12/20/13 m.michael.musatov@gmail.com
12/20/13 m.michael.musatov@gmail.com
12/20/13 m.michael.musatov@gmail.com
12/20/13 m.michael.musatov@gmail.com
12/20/13 m.michael.musatov@gmail.com
12/20/13 m.michael.musatov@gmail.com
12/20/13 m.michael.musatov@gmail.com