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Nov 3, 2010 at 15:15 vote accept fred goodman
Nov 3, 2010 at 15:08 vote accept fred goodman
Nov 3, 2010 at 15:09
Oct 28, 2010 at 22:11 answer added fred goodman timeline score: 0
Oct 27, 2010 at 22:02 comment added Kevin Buzzard Yeah, I think so. One can solve a general linear recurrence over an alg closed field: the general solution is that $\omega_a$ is a sum of things of the form $h(a).x^a$ with $h$ a polynomial and $x\in F$. Now your second assumption implies $h(a)^2=h(2a)$ for all integers $a\geq0$, but you are in char 2 so $h(2a)=h(0)=c$, the constant term, and so $h(a)$ is the unique (again as you're in char 2) square root of $c$ for all $a$, so $h$ may as well be replaced by a constant function $c$ satisfying $c=c^2$, and $c=1$ is the only interesting solution, giving you the solution you already spotted.
Oct 27, 2010 at 21:31 history asked fred goodman CC BY-SA 2.5