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Prove that $[2][x]^{68}+[14][x]^{22}+[63][x]^2+[2]$ does not have any roots in $\mathbb{z}_{11}$.

First I simplified the coefficients: $14 \equiv 3 \pmod{11}$ and $63 \equiv 8 \pmod{11}$

Next we can simplify the larger exponents using Fermat's little theorem: $x^{68} = x^8$ and $x^{22} = x^2$.

Then, in $\mathbb{z}_{11}[x]$ we get

$2x^8+3x^2+8x^2+2$

which can be simplified to

$2x^8 + 2$

I'm stuck at this point. How do I show that there are no roots to this polynomial?

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  • $\begingroup$ I didn't lie, multiply through by $6$ $\endgroup$ – mdave16 Mar 29 '17 at 22:33
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hint: since we may rule out $x=0$ and $x^{10}=1$ in $Z_{11}$ the result of your simplification can also be written as $$2x^{-2}(x^{10}+x^2)=2x^{-2}(1+x^2)$$

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Hint:

This is equivalent to $x^8\equiv -1\equiv10\mod 11$. This implies $10$ is a square mod. $11$. Determined the list of squares in $\mathbf Z/11\mathbf Z=\bigl\{[0], \pm[1],\pm[2],\pm[3],\pm[4],\pm[5]\bigr\}$.

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