Does $x^3-x+3$ have any roots in $\mathbb{F}_5$? I don't think it does but the only way I know to check is by trial and error and there's only so many factors I can try before my hand cramps up.
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3This is small enough you don’t even need to write anything. Just calculate the numbers in your head... – Clayton Mar 29 '18 at 17:36
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@Clayton One doesn't even need a calculation. Use the "cubing lemma" with $p = 5$ and $k = 3$. Then every number is a cube. – GNUSupporter 8964民主女神 地下教會 Mar 29 '18 at 17:40
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1@GNUSupporter But this cube ($x^3)$ need not coincide with $x-3$ – Peter Mar 29 '18 at 17:46
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2You don't have to check the factors with polynomial long division, you only have to insert the numbers $0,1,2,3,4$ and look whether the value of the polynomial is divisible by $5$ – Peter Mar 29 '18 at 17:49
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@Peter Sorry for my careless mistake. – GNUSupporter 8964民主女神 地下教會 Mar 29 '18 at 18:02
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1@Peter Trying $0,1,2,3,4$ seems awfully difficult. Better to try $-2,-1,0,1,2$. – David C. Ullrich Mar 29 '18 at 18:23
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Note $$x^3-x=x\left(x^2-1\right) = (x-1) x (x+1)$$ hence $-1, 0, 1$ will not work as you need $x^3-x=-3 \equiv 2 \pmod 5$, check the other values $\pm 2$?
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