# sums of factors in GF(2)

franktaw at netscape.net franktaw at netscape.net
Wed Aug 2 15:34:30 CEST 2006

```That is indeed impossible.  If p(x) in GF(2)[x] is irreducible (and not
x+1), then in particular p(1) = 1, and then p(1) + 1 = 0, so p(x)+x is
divisible by x+1.

-----Original Message-----
From: Antti Karttunen <antti.karttunen at gmail.com>

Marc LeBrun wrote:

> Recently I've been trying to exhort the Sequence Phanatiques to >
compute analogs of the sum-of-prime-factors (with and without >
multiplicity) in other arithmetics, such as Gaussian integers, GF(2), >
etc.
>
Don't worry, I will do the GF(2)[X] -thing, in my due time. (This is
easy to do
by using the existing the
http://www.research.att.com/~njas/sequences/a091247.scm.txt
Analogues for http://www.research.att.com/~njas/sequences/A000203 and
similar sequences
as well...)

> I was wondering, specifically about GF(2), summing (ie XORing) the >
prime factors of N with multiplicity:
>
> Noting that only the square-free part of N matters, since the square
> parts sum to 0...
>
> A. Aside from the perfect squares (eg 5) are there any other N that >
sum to 0? Can they be characterized?
>
Antti:

If we could just find from
http://www.research.att.com/~njas/sequences/A014580
two successive terms, with first A014580(k) = 1 mod 4, and the second
A014580(k+1) = A014580(k)+2,
then by xoring them we could get 2, and the product A014580(k) x
A014580(k+1) x 2
(where x would be GF(2)[X] multiplication:
http://www.research.att.com/~njas/sequences/A048720 )
would satisfy the condition.
However, skimming cursorily over the first 100 terms of
http://www.research.att.com/~njas/sequences/A058943
I didn't find such a pair. (Maybe there's a reason for that? GF(2)[X]
secrets elide me for a monent...)

Yours,

Antti

```