[seqfan] Re: Pairs Occurring Only Once Among # Of Divisors
maximilian.hasler at gmail.com
Fri Jun 26 04:46:01 CEST 2009
>> On behalf of the Quetau pairs A161640 invented in
> It is also easy to prove 1024:(...)
> I'm guessing that a similar proof can be worked out for
> n = 2^(p-1) where p is prime. This proof was for p=11.
It is also easy to prove 2^12 = 4096.
d(2^12) = 13
d(4097) = 4
If n has 13 divisors, then p^12 is the only possible form, with p prime.
If n+1 has 4 divisors, then it is q^3 or qq' for distinct primes q and q'.
However, n+1 = p^12+1 = (p^4+1)(p^8-p^4+1).
For p>2, this is the product of the even number p^4+1 >= 82
and the odd number p^8-p^4+1 >= 3^8-3^4+1,
which can't be of the form q^3 or qq'.
So the only solution is p=2.
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