On 10/24 Jose Ramon Brox wrote: Suposse n and n+1 get together all the primes below n. By Bertrand s Postulate, there s always a prime p between k and 2k. 1)

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, Oct 29, 2005

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On 10/24 Jose Ramon Brox wrote:

Suposse n and n+1 get together all the primes below n.

By Bertrand's Postulate, there's always a prime p between k and 2k.

1) If n is even, between n/2 and n there's a prime p.
1.a) If p>n/2, then p can't divide neither n nor n+1, because gcd
(n,n/2+k),
gcd(n+1,n/2+k)<(n/2+k).
1.b) If p=n/2, then n = 2*p, but then n+1 must be
3*5*...*nearest_prime_to_n
(except p)
and then it is bigger than n+1.
2) If n is odd, then n+1 is even. Between (n+1)/2 and n+1 there's a
prime p.
2.a) If p>(n+1)/2 then p can't divide neither n nor n+1.
2.b) If p=(n+1)/2 then n+1 = 2*p, but then n must be
3*5*...*nearest_prime_to_n
(except
p), and this is only affordable if p=3 and n=5, otherwise n is
bigger than n+1.
---------------------------

Thanks for the idea, Jose. I like the approach you used, but I do
think there are a few issues in your proposed solution. I have been
trying to modify your approach to make it work, but as yet no
success. Here are what I see as some of the issues:

A. I am looking for a solution that also includes n as a factor if
n is prime, not just the primes below n.

B. In statement 2.a) you say "If p>(n+1)/2 then p can't divide
neither n nor n+1." Not true given my comments in "A" above. If
p=n, then of course p divides n.

C. According to Mathworld, Bertrand's postulate states that there
is always a prime p such that k<p<2k. Therefore, the equalities
(p=n/2 in statements 1.b) and (p=(n+1)/2 in statement 2.b) are
incorrect.

D. It is not clear to me how you arrive at your final statement
"and this is only affordable if p=3 and n=5, otherwise n is bigger
than n+1." Why is it only affordable if p=3 and n=5?

Thanks for your suggestions. I hope there is a solution lurking
somewhere in here. Regards, Bill

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