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The answer paper says we can do it like this $(x^2+x+1)(x^3+x+1)$, but I dont know how they got that. Can you please write a step by step tutorial?:D Thank you! (Since $0$ and $1$ are not roots, I can’t even start to factorize)

Math1000
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Elekhey
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    The number of (irreducible) polynomials in $\Bbb F_2[x]$ of degree lower or equal to $n$ is finite, so this possibility is always at hand. –  Dec 09 '19 at 23:32
  • Note that the only way it can be factored is into degree 2 and 3, as it's quite easy to prove it has no rational roots. 4+1 or 3+1+1 or 2+2+1 for example all include linear roots – Gabe Dec 09 '19 at 23:55
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    This factorization is essentially also true over the integers: https://math.stackexchange.com/questions/2286149/prove-that-n5n41-is-composite-for-n1/2287249?noredirect=1#comment4707391_2287249 – Gunnar Sveinsson Dec 10 '19 at 09:31

2 Answers2

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If it can be factored, it is factored as the product of irreducible polynomials of degree $2$ and $3$ respectively since it has no roots. So we may try to see whether it is divisible by an irreducible quadratic polynomial, i.e. a quadratic polynomial with no root. Over $\mathbf Z_2$, the single quadratic irreducible polynomial is $x^2+x+1$, since the other quadratic polynomials are $$x^2+x=x(x+1)\quad\text{ and }\quad x^2+1=(x+1)^2.$$ It happens the quotient is $x^3+x+1$, and it is irreducible since it has no root either.

Bernard
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Hint: Let $f=x^5+x^4+1$. Then:

  • $\gcd(f,x^2-x)$ is a possible factor of $f$ of degree at most $1$

  • $\gcd(f,x^4-x)$ is a possible factor of $f$ of degree at most $2$

  • $\gcd(f,x^8-x)$ is a possible factor of $f$ of degree at most $3$

  • $\gcd(f,x^{32}-x)=f$ iff $f$ is irreducible mod $2$

All gcds can be easily found with the Euclidean algorithm.

Solution:

  • $\gcd(f,x^2-x)=1$

  • $\gcd(f,x^4-x)=x^2 + x + 1$

  • $\gcd(f,x^8-x)=x^3 + x + 1$

  • $f=(x^2 + x + 1)(x^3 + x + 1)$

The two factors are irreducible mod $2$ because they have no roots.

lhf
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