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The following problem is from an exam Coding and cryptography of 2017, where $K\in \lbrace 0,1\rbrace$ so that we work with binary codes.

Let $C \subseteq K^{60}$ be the cyclic linear code with generator polynomial $g(x) = 1 + x^6$. How many cyclic linear codes $C' \subseteq C$ are there? (Hint: how does the generator polynomial of $C'$ relate to $g(x)$?)

In the book of Hankerson they we can factorize $1+x^{60}$ as follows. Since $x=2^2\cdot 15 = 4\cdot 15$, then $1+x^{60} = (1+x^{15})^4$. Next I want to know $z$, the value of different irreducible polynomials if $1+x^{15}$ is factorized. From this it should be doable to calculate the number of cyclic codes with a $g(x)$ in it.

Also, there is a procedure to factorize $x^{15}+1$, and this has do to with idempotent polynomials and gcd, as in Hankerson chapter 4, but that calculation would be too long for only $4$ of $90$ points.

Does anyone know an argument? Thanks in advance.

Rocco van Vreumingen
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    Factorization of $1+x^{15}$ is given here. BUT, answering the exercise is rather about factoring $1+x^6$, which is a simpler task... – Jyrki Lahtonen Mar 27 '18 at 05:11
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    Have you thought about the hint? What is your answer? – leonbloy Mar 27 '18 at 11:41
  • Yes I have already thought about this idea, even before Jyrki gave it. But I did not know that this was meant as a hint, because there are also people who don't know the answer but only a part of it and write that part down. – Rocco van Vreumingen Mar 28 '18 at 09:42
  • However I already have an answer: the algorithm in Hankerson give $32=2^5$ divisors for $1+x^{15}$, so $1+x^{15}$ must be of the form $g_1(x)g_2(x)g_3(x)g_4(x)g_5(x)$ where all $g_i$ are irreducible. So then $1+x^{60} = \prod_i g_i^4$, and we know $1+x^6$ is part of it, so this gives that $1+x^{60} = (1+x^6)g_1(x)^2g_2(x)^2g_3(x)^4g_4(x)^4g_5(x)^4$ so the answer is $3\cdot 3 \cdot 5^3 = 1125$ – Rocco van Vreumingen Mar 28 '18 at 09:47

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