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How many elements $x$ in the field $\mathbb{ Z}_{11}$ satisfy the equation $x^{12} - x^{10} = 2$?

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3 Answers 3

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Hint: every nonzero element $x \in \mathbb{Z}/11\mathbb{Z}$ satisfies $x^{10} = 1$.

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    $\begingroup$ Somebody has downvoted all three answers to this question. Would they please explain their reasoning? $\endgroup$ Aug 12, 2013 at 18:12
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Hint: You just have to solve the congruence

$$x^{12}-x^{10}\equiv2\pmod{11}$$

Then apply the Fermat's little theorem.

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    $\begingroup$ Instead of putting the parenthesis yourself, use \pmod{n} to get $\pmod{n}$. The braces are used to avoid, say $\pmod 123$ in contrast to $\pmod{123}$. $\endgroup$
    – Pedro
    Aug 11, 2013 at 2:46
  • $\begingroup$ Alright thanks for the tip, will do for the future! $\endgroup$
    – user67258
    Aug 11, 2013 at 2:48
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Obviously $x \equiv 0 \pmod{11}$ is not a solution, therefore $x$ is non-zero in $\mathbb{Z}_{11}$ and we can factorize this way:

$x^{12}-x^{10} \equiv 2 \pmod{11} \implies x^{10} ( x^{2}-1) \equiv2 \pmod{11}$ but $x^{10}\equiv 1 \pmod{11}$ for every non-zero $x \in \mathbb{Z}_{11}$ by Fermat's little theorem, or more generally using Euler's totient function $\varphi(n)$. That implies:

$x^2 - 1 \equiv 2 \pmod{11} \implies x^2\equiv 3 \pmod{11} \implies x \equiv 5,6 \pmod{11}$

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