[seqfan] Re: Pairs Occurring Only Once Among # Of Divisors
Hagen von EItzen
math at von-eitzen.de
Thu Jun 25 22:42:31 CEST 2009
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.
>
>
The first conjecture
n = 35, tau(n) = 4, tau(n+1) = 9
is true:
We must have n = p^3 or pq, n+1 = r^8 or r^2 s^2
a) n even:
Then n = 2^3 or n = 2q with odd q
and n+1 = r^8 or (rs)^2 with odd r,s.
But odd squares are 1 (mod 8), hence n = 0 (mod 8), i.e. n = 2^3. But 9
is not r^8 and not (rs)^2.
Hence n cannot be even.
b) n odd:
Then n+1 = 2^8 or 4s^2 and n=p^3 or pq (with p,q,s odd).
255 = 3*5*17, hence n+1 = 2^8 can be excluded.
Then n = 4s^2 - 1 = (2s1)(2s-1)
If n = p^3, this implies
- either 2s-1 = 1 => s=1 => contradiction
- or 2s-1 = p, 2s+1 = p^2 => p^2 = p+2 => p in {2,-1} => contradiction.
Remains the case
n = p*q = (2s+1)*(2s-1)
Wlog. p>q.
Then p=2s+1 and q=2s-1 are twin primes, hence 2s is a multiple of 6, s a
multiple of 3.
Hence s=3, leading to n= 35.
Hagen
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