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

[Hagen von EItzen]

Richard Mathar schrieb:
> I've updated the script in
> http://www.strw.leidenuniv.nl/~mathar/progs/a161460.pdf
> with proofs that have been posted here and with proofs on 728 and 57121.
>
> Richard Mathar
>   
I somehow feel like my proof methods could be mechanized, at least for 
cases with tau(n+1) = 3 (or possibly odd prime), but here ya go with 
another explicit case:

n= 528, tau(n)=20, tau(n+1)=3:
n+1 is the square of a prime and surely != 4, hence n is even.
n is one of 2^19 (but then tau(n+1)=4), 2^9*p, 2^4*p^3, 2^4*p*q, 
2^3*p^4, 2*p^9, 2*p^4*q.
Since n+1 is an odd prime square r^2, n = 0 (mod 8),
thus leaving cases 2^9*p, 2^4*p^3, 2^4*p*q, 2^3*p^4.
In n = (r-1)(r+1), both factors are even and are coprime apart from that 
and exactly one is 0 mod 4.
* If n = 2^9*p these factors must be 2^8 and 2p, which implies r=257 
(255 is composite) and hence p = (r+1)/2=129 = 3*43 -- contradiction.
* If n = 2^4*p^3, r+-1 must be 8 and 2p^3, which is impossible.
* If n = 2^3*p^4, r+-1 must be 4 and 2p^4, whic is impossible
* If n = 2^4*p*q, r+-1 must be 8 and 2pq or (wlog.) 8p and 2q. The first 
case is quickly ruled out by inspection, IN the second case, r != 3 (as 
n != 8), hence one of r+-1 is a multiple of 3, i.e. p=3 or q=3. If q=3 
then r = 6+-1, n = 24 or n = 48. but then tau(n) != 20.
If p = 3, then r=24+-1, hence r=23 and n = 528 as was to be shown


Hagen