p^(2k+1) for any prime works, since (x+1) | (x^n + 1) when n is odd. -----Original Message----- From: math-fun-admin@mailman.xmission.com [mailto:math-fun-admin@mailman.xmission.com]On Behalf Of Gershon Bialer Sent: Wednesday, January 22, 2003 12:51 To: math-fun@mailman.xmission.com Subject: Re: [math-fun] Re: your mail Offhand, I can think of lots of numbers that satisfy the conjecture. Consider a mersenne prime p=2^m-1 for some integer m. Let n=p^(2k+1), and p+1|n+1 since p+1=2^m and n(mod 2^m)=(-1)^(2k+1)+1=0. Gershon Bialer
On Wed, 22 Jan 2003, Mr. Nayandeep Deka Baruah wrote:
Dear Professors Guy and Borwein,
I would like to know from you whether the following result is still a conjecture or has been proved by somebody.
There exists a composite integer n such that for each prime divisor p of n (p+1)|(n+1).
If it is true then what is the smallest such number? Are such numbers are infinitely many?
I would be extremely grateful for your help.
With best regards,
Nayandeep Deka Baruah Dept. of Math. Sciences, Tezpur University Napaam-784028 Assam, INDIA.
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