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This is homework exercise: $$P=t^{1024} + t + 1 , R = \mathbf{F}_{2}[t] \Rightarrow P \ \text{reducible in R}$$

I wanted to show this analogous to how a book shows it (book shows it with other numbers and field): $a^{1024}= a+1$ has solutions in $R/PR$, then calculate $a^{2^{20}}, a^{2^{1024}}$ in $R/PR$ and by using gcd conclude that P is reducible over R.

My problem is that the numbers are so big that I can't split up and show that there are solutions or calculate $a^{2^{20}}, a^{2^{1024}}$ in $R/PR$. But there must be an easy way since it is homework. So how to show that $a^{1024} = a+1$ is solvable in $R/PR$ and how to compute $a^{2^{20}}, a^{2^{1024}}$ in $R/PR$? Thanks for all input.

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    By brute force: $t^4+t+1$ divides $P.$2011-12-22
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    Thanks for your comment. Is there no way to show it the way the book does it? Because I can't brute force in the exam.2011-12-22
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    $a^{1024}=a+1$ is automatically solvable in $R/PR$, and a root is the residue of $t$. Seeing why this is true is a good test of your understanding of the definition of $R/PR$. If it is not clear why $a^{1024}=a+1$ has a root in $R/PR$, spend some time with the definition of $R/PR$.2011-12-22
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    @Ben: i find patronizing comment with zero math content not specialy usefull.2011-12-22
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    @evgeniamerkulova: Tashi can judge for self but I certainly wasn't feeling patronizing when I wrote it. Was trying to give good advice. If not clear why $a^{1024}=a+1$ is solvable in $R/PR$, definition of $R/PR$ is right place to start.2012-01-10

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