# [seqfan] Re: Pairs Occurring Only Once Among # Of Divisors

T. D. Noe noe at sspectra.com
Fri Jun 26 01:13:37 CEST 2009

```At 10:42 PM +0200 6/25/09, Hagen von EItzen wrote:
>Richard Mathar schrieb:
>> On behalf of the Quetau pairs A161640 invented in
>>
>> http://list.seqfan.eu/pipermail/seqfan/2009-June/001652.html
>>
>> I started some sort of explicit proofs of these in
>> http://www.strw.leidenuniv.nl/~mathar/progs/a161460.pdf .
>> The interesting part starts at page 116. This is an unfunded
>> project and will require some Jovian years to complete.
>>
>>

It is also easy to prove 1024:

d(1024)=11
d(1025)=6

if n has 11 divisors, then s^10 is the only possible form with s prime.

if n+1 has 6 divisors, then it is p^5 or pq^2 for distinct primes p and q.

Suppose n is odd.  Then n+1 is even, implying that n+1 is either 2^5, 2q^2,
or 4p.  Note that n+1 = s^10+1 = (s^2+1)(s^8-s^6+s^4-s^2+1).  Because s is
odd, the s^2+1 factor equals 2f for some odd f.  Hence, the form of n+1 is
not 2^5 or 4p.  It cannot be 2q^2 because that would require (s^2+1)/2 =
s^8-s^6+s^4-s^2+1, an equation whose integer roots are s = -1 and 1.

Suppose n is even.  This implies s^10 is even, which implies s=2.  This
gives us the only solution, n=1024.

I'm guessing that a similar proof can be worked out for n = 2^(p-1) where p
is prime.  This proof was for p=11.

Tony

```