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Am I doing this problem right?

Use Fermat's Little Theorem to find all the roots of the following polynomials in $\mathbb{Z}_{7}[x]$: $2x^{74}-x^{55}+2x+6$

If I know that Fermat's Little Theorem states

$a^{p-1} \equiv1 \pmod{p} $

Therefore if the if we are using $\mathbb{z}_{7}[x]$ we can use $a^6 \equiv1 \pmod {7}$

Which will allow for: $$2(x^{74})-(x^{55})+2x+6=$$ $$2\Big(x^{(6*12)=72}\equiv 1 \pmod{7}\Big)x^2-\Big(x^{(6*9)=54} \equiv 1 \pmod 7\Big)x+2x+6=$$ $$2x^2-x+2x+6=2x^2+x+6$$

Am I doing this right ? Any feedback would be greatly appreciated

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  • $\begingroup$ Looks great so far. To see how that final quadratic factors, you might want to write the middle term as $8x$ instead of $x$. $\endgroup$
    – G Tony Jacobs
    Commented Nov 10, 2017 at 15:13
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    $\begingroup$ You need also to consider $x=0$ to eliminate the possibility. $\endgroup$
    – Mark Bennet
    Commented Nov 10, 2017 at 15:14
  • $\begingroup$ That's a good point. FLT doesn't apply if $x\equiv 0$, but that's easy to rule out, in this case. $\endgroup$
    – G Tony Jacobs
    Commented Nov 10, 2017 at 15:15

1 Answer 1

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It's obvious that $0$ is not a root. If $a\ne0$ is a root, then $$ a^{74}=a^{6\cdot12+2}=a^2, \qquad a^{55}=a^{6\cdot9+1}=a $$ and therefore $2a^2-a+2a+6=0$ or $2a^2+a-1=0$; the quadratic formula gives…

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