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

Hagen von EItzen math at von-eitzen.de
Fri Jun 26 08:12:17 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.
>
> Richard Mathar
>
>
> _______________________________________________
>
> Seqfan Mailing list - http://list.seqfan.eu/
>
>
>
The remaining n < 100 work out, too:

(Below, all prime variables stand for odd primes)

I) tau(n) = 6, tau(n+1) = 7:
a) n even.
Then n = 2^5 or n = 2p^2 or n = 4p.
tau(n+1) is odd, hence n+1 must be square and = 1 (mod 8).
This matches only n=2^5, which is no solution, tau(33) != 7
b) n odd.
Then n+1 is even and must be 2^6
Hence n = 63.

II) tau(n) = 7, tau(n+1) = 4:
a) n odd:
n = p^6 and n+1 = 2^3 (but 7 != p^6) or n+1 = 2q.
But (p^6+1)/2 = (p^2+1)/2 * (p^4-p^2+1) is a nontrivial factorization of q.
b) n even:
Then n = 2^6 as was to be shown.

III) tau(n) = 10, tau(n+1) = 5:
a) n odd.
Then n+1 = 2^4, but tau(15) != 10
b) n even.
Then n = 2^9 (but tau(513) != 5) or n = 2*p^4 or n = 16*p.
Note that n+1= q^4 is an odd square, is 1 (mod 8), hence n = 0 (mod 8),
which leaves only n = 16p.
But q^4 - 1 = (q-1)*(q+1)*(q^2+1). All three factors are even, hence
exactly one is divisivble by 4.
Also, exactly one is divisible by p. If not p|q^2+1, we conclude
Hence q- 1=2, q+1 = 4, q^2+1 = 2p, i.e. n = 80.

IV) tau(n) = 6, tau(n+1) = 9:
a) n even.
n = 2^5 (but tau(33) != 9) or n = 2p^2 or n = 4p and once again n+1 is
an odd square, hence n must be 0 (mod 8) -- contradiction.
b) n odd.
n = p^5 or n = p*q^2.
n+1 = 2^8 (but tau(255) != 6) or n+1 = 4r^2.
Then p q^2 = (2r+1)(2r-1).
At most one of 2r+1, 2r-1 is divisible by q, hence
1) either p = 2r-1, q^2 = 2r+1
Since r is prime, r = +-1 (mod 6), hence 2r-1 = 1 or 3 (mod 6). Since p
is prime, only p = 2r-1 = 1 (mod 6) remains.
But then q^2 = 3 mod 6 -- contradiction.
2) or p = 2r+1, q^2 = 2r-1
Then p = q^2 + 2. If q=3, we obtain p = 11 and n = 99.
Otherwise, this implies p=0 (mod 3) and q^2=1 -- contradiction
Hence we must have n = 99.

Hagen

```