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How would I show that $p(x)=x^5+x^2+1$ is an irreducible polynomial over $\Bbb Z_2=\{0,1\}$.

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closed as off-topic by Travis, Najib Idrissi, Christopher, Dario, TravisJ May 18 '15 at 12:59

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  • $\begingroup$ please take this off hold. $\endgroup$ – Maths student May 18 '15 at 19:11
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Hints. It has no factors of degree one (why?). Then check whether your polynomial is divisible by the only irreducible polynomial of degree two over $\mathbb F_2$: $X^2+X+1$.

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    $\begingroup$ One can see this nicely without carrying out long division: If $x^2 + x + 1$ divided $p(x)$, it would also divide $p(x) - (x^2 + x + 1) = x^5 + x = x(x^4 + 1) = x(x^2 + 1)^2 = x(x + 1)^4$, which is a product of five linear factors, a contradiction. $\endgroup$ – Travis May 18 '15 at 11:01
  • $\begingroup$ @Travis Nice trick! $\endgroup$ – user26857 May 18 '15 at 11:05
  • $\begingroup$ One can also derive a contradiction by setting $x^5 + x^2 + 1 = (x^2 + x + 1)(ax^3 + bx^2 + cx + d)$ and setting up a linear system of the coefficients, but Travis's method is much, much cleaner. $\endgroup$ – Fargle May 18 '15 at 11:07
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    $\begingroup$ Thank you. I like user26857's method the most because it helps me with the next part of the question as well. :) $\endgroup$ – Maths student May 18 '15 at 11:09
  • $\begingroup$ @user26857 Yes. $a = d = 1$, and then one derives both $b+c = 1$ and $b+c = 0$. Though I suppose it is trivial that $a = d = 1$, so I guess I only really needed to specify the middle two coefficients. $\endgroup$ – Fargle May 18 '15 at 11:10

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