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

Hagen von Eitzen hagen at von-eitzen.de
Thu Jun 25 22:55:47 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 second conjecture
n = 48, tau(n) = 10, tau(n+1) = 3
is also true:

a) n odd:
Then n+1 = 2^2, but tau(3) = 2

b) n even:
Then n = 2 * p^4 or n = 16 * p (p odd)
and n+1 = q^2 (q odd).
Since q^2 = 1 (mod 8), only n = 16p remains.
Then 16p = (q+1)(q-1) implies that q-1, q+1 are complementar divisors of
16p.
One of them is not a multiple of p, hence among 1, 2, 4, 8, 16 and the
other is accordingly 16p, 8p, 4p, 2p or p.
8p - 2 and 4p - 4 are >2, hence divisor pairs (2,8p) and (4,4p) can be
ruled out.
Since q+1 and q-1 have the same parity, (1,16p) and (16,p) can be ruled
out as well.
We are left with (8,2p). Then q = 8-1 = 7 (as 8+1 = 9 is not prime) and
n = 48 as was to be shown.

Gee, I sohould not look too long at the list of conjectures.
One might get caught in manually proving them all ...

Hagen

```