[seqfan] Re: Pairs Occurring Only Once Among # Of Divisors
Hagen von EItzen
math at von-eitzen.de
Sat Jun 27 12:09:46 CEST 2009
Hagen von EItzen schrieb:
> 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:
>
>
Still another one, I hope collecting all thos tiny results does not get
boring:
n = 575, tau(n) = 6, tau(n+1) = 21:
Then n = p^5 or n = p^2*q and n+1 = r^20 or n+1 = r^6*s^2.
If n+1=r^20, then n = r^20-1 = (r-1)(r+1)(r^2 +
1)(r^4-r^3+r^2-r-1)(r^4+r^3+r^2+r+^)(r^8-r^6+r^4-r ^2+1).
The case r=2 is eliminated explicitely: tau(2^20-1)=48.
Hence n can be written as product of 6 factors >=2, contradicting the
prime signature(s) of n.
Hence n+1 = r^6*s^2. If n+1 is odd, then as usual n = 0 (mod 8), hence n
= 2^5. But tau(33) = 4.
Therefore, n is odd and can be written as n = (r^3*s - 1)(r^3*s + 1),
i.e. as product of adjacent and hence coprime odd numbers. This
eliminates the case n = p^5 (|p^k-p^(5-k)| >= (p-1)p^2 >2) and
enforces |p^2 - q| = 2.
q=3 is impossible and the case p=3 (with q=7 or q=11) can be eliminated
eplicitly: tau(63+1)=7, tau(99+1)=9.
Therefore p^2 = 1 (mod 24) implies q = -1 (mod 24) and q = p^2 - 2.
But then p^2 = r^3*s +1 and r^3*s must be 0 (mod 8), hence r = 2.
Thus we arrive at the conditions: p^2 = 8s + 1, q = 8s - 1.
Exactly one of s, 8s+1, 8s-1 must be 0 (mod 3), hence one of p,q,s is 3.
The only valid possibility is s=3, p=5, q = 23, leading to n = 575.
Hagen
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