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How do I split $x^2-5 $ in $\mathbb{Z}/5\mathbb{Z}$ ? Since $0$ is a root I have $x $ as linear factor . How can I find the other linear factor ?

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closed as off-topic by Servaes, cmk, ronno, Adrian Keister, ThorWittich Jul 22 at 20:37

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    $\begingroup$ $x^2=x\times x$ $\endgroup$ – J. W. Tanner Jul 21 at 20:21
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    $\begingroup$ -1 This really shows very little effort; there are only $4$ other candidates for the roots, and none of them are roots. Or you could have tried polynomial long division. Or simplified $x^2-5$ to $x^2$. $\endgroup$ – Servaes Jul 22 at 11:30
  • $\begingroup$ @Servaes: I'd say that this shows more effort than a lot more recent questions. $\endgroup$ – user21820 Jul 29 at 15:25
  • $\begingroup$ @user21820 Regardless of the quality of other questions, I find this one deeply subpar. But feel free to link some more questions to close. $\endgroup$ – Servaes Jul 29 at 16:56
  • $\begingroup$ @Servaes: Sure. You can find many here (especially in the middle and bottom). $\endgroup$ – user21820 Jul 29 at 17:42
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Note that $x^2-5 = x^2$, which is already factorised.

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Hint: Vieta's formulas still work. The sum of the two roots equals the (negative of the) coefficient of the first degree term.

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