How to find a complex root of $x^{2021}=x^{2020}+1$ while this complex root also satisfies a quadratic equation with integer coefficients? I have no previous experience in solving complex equations so just have no clue on this kind of question... Is there any useful pattern on this?
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3Obviously, the first thing to do is to look at what happens to $x^{N+1}-x^N-1$ when $N=2020$ is replaced by smaller, more manageable values. Give it a try. – Ewan Delanoy Apr 04 '20 at 18:05
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1Hint: $x^5 - x^4 - 1 = (x^2 - x + 1) ( x^3 - x - 1)$. – Calvin Lin Apr 04 '20 at 18:13
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Hint: $x^5 - x^4 - 1 = (x^2 - x + 1) ( x^3 - x - 1)$.
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Hint: $x^3 +1 = (x^2 - x + 1 ) ( x + 1)$.
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Hence, $ x^{6n+k} \equiv x^{k} \pmod{x^2-x+1}$
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Hence $ x^{2021} - x^{2020} - 1 \equiv x^5 - x^4 - 1 \equiv 0 \pmod{x^2-x+1}$.
Calvin Lin
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