[seqfan] Re: A property of 163

franktaw at netscape.net franktaw at netscape.net
Fri Jul 10 16:48:49 CEST 2009

Ah, now it becomes clear: the 4 is a typo; it should be 3.

I've been wondering recently about Z[i,sqrt(n)].  Which of these
domains has unique factorization?

Incidentally, Z[i,sqrt(2)] is an example of a finite dimensional
extension to the rationals, in which every rational prime has a
non-trivial factorization.  Primes == 3 (mod 8) factor in Z[sqrt(-2)],
p == 5 (mod 8) factor in Z[i], p == 7 (mod 8) factor in Z[sqrt(2)], and
p == 1 (mod 8) factor in all 3 (as does 2).  It's pretty clear that no
quadratic extension of the rationals can have this property; I have no
idea how to determine whether any cubic extension does.  (The
quaternions also have this property (see four squares theorem), but
that isn't an integral domain, and anyhow it's also 4 dimensional.)

Another note: for any (square-free) n with absolute value > 1,
Z[i,sqrt(n)] has non-trivial elements.  If n == 1 (mod 4), it has (1 +
sqrt(n))/2; if n == 3 (mod 4), it has (1 + sqrt(-n))/2; and if n == 2
(mod 4), it has (sqrt(n) + sqrt(-n))/2.

-----Original Message-----
From: drew at math.mit.edu

The unique property of 163 noted below is correct. The proposed
sequence is
missing 3 and is effectively already in the OEIS, see A014602.

Regards,

Drew

On Jul 10 2009, Tanya Khovanova wrote:

>Dear SeqFans,
>
> I received the following submission for my number gossip page
> (numbergossip.com) from Anand Deopurkar:
>
> "A unique property of 163: It is the largest number n such that the
> integers in the imaginary quadratic extension Q(\sqrt -n) have the
unique
> factorization property."
>
>Can some confirm this?
>
>He also sent a sequence which is not in the database:
>
> "Integers in the following imaginary quadratic fields Q(\sqrt -n)
have
> the unique factorization property: n = 1,2,4,7,11,19,43,67,163. So
you
> could add this as a rare property for those integers as well."
>