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As the title suggests, I want to describe all $f \in \mathbb{F}_2[x]$ divisible by $x^2 +1$. I am sure it's easy but I could't do it for like an hour so I wrote this question.

What I know.

  1. It's easy for $x+1$ instead of $x^2+1$. In this case the answer is simply the polynomials with even number of nonzero coefficients.

  2. The remainder is either $x+1$ or zero, therefore there is a bijection between polynomials I need to describe and polynomials I don't need which takes $f \mapsto f + (x+1)$ (the inverse map is the same).

Bill Dubuque
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Invincible
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    Write $f(x)=g(x^2)+xh(x^2)$ for some $g,h\in\mathbb{F}_2[x]$. Note that $f(x)=(x^2+1)q(x)$ iff replacing $x^2=1$ you get $0$. So, this is the same as $g(1)=h(1)=0$. You can replace this condition by your condition (1) for $g$ and for $h$. So, the even part and the odd part of $f$ must both have an even number of non-zero coefficients. – plop Nov 30 '21 at 23:27
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    Note that $x^2+1=(x+1)^2$ modulo $2$. – markvs Nov 30 '21 at 23:30
  • By the way, regarding your observation (2). If $f(x)=x(x^2+1)+1$ the remainder after division by $x^2+1$ is neither $x+1$ nor $0$. It is $1$. You can also see the possible remainders as the possible values of $g(x^2)+xh(x^2)$ when you replace $x^2=1$. The $g(1)$ and the $h(1)$ can take each of the values $0,1$. – plop Nov 30 '21 at 23:35
  • @Boxwood, it is exactly how I know the bijection. – Invincible Dec 01 '21 at 11:57

1 Answers1

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First, show that if $f(x)=f_0+f_1x+\dots+f_nx^n$, then $$ f(x)\equiv (f_0+f_2+\dots)+x(f_1+f_3+\dots)\pmod{x^2+1} $$ Conclude that $f(x)$ is a multiple of $x^2+1$ if and only if the sum of the even index coefficients is $0$, as well as the sum of the odd-index coefficients.

Mike Earnest
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    Or apply the double root test to infer $,(x!-!1)^2\mid f(x)\iff f(1)=0= f'(1),,$ and note $\bmod 2!:\ f'(1)\equiv f_1 + f_3 + f_5 + \cdots $ – Bill Dubuque Dec 01 '21 at 02:09
  • Please have the courtesy to cite a comment when you duplicate it in an answer, esp. when it 's very old (a couple hours here). – Bill Dubuque Dec 01 '21 at 02:11
  • @BillDubuque I wrote this answer without looking at that comment. – Mike Earnest Dec 01 '21 at 02:13
  • Given that it is the first comment on the answer, that seems quite difficult to do even when trying (I cannot force it outside my field of vision while reading the question). Please do read the comments since often they contain pertinent information necessary to formulate a good answer. Common site courtesy is to make such an answer CW and cite the comment. – Bill Dubuque Dec 01 '21 at 02:22
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    Meh! Nobody is inventing anything new here and this site is a collaboration. Only the idea deserves reverence, if it works, and not the people. I certainly don't need any acknowledgement. – plop Dec 01 '21 at 03:26