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Number theory related question.

Give all answers to:

$$3x^{10}\equiv 10x^3 \pmod{13}$$

$0$ is obvious but I can't see a good way to draw out $12$.

I've got this so far:

Rearrange to $3x^{10}-10x^3\equiv 0 \pmod{13}$

Factor out $x^3$, to give:

$x^3(3x^7-10)\equiv 0\pmod{13}$

$0$ works because of $x^3$ term

I'm still looking for:

$(3x^7-10)\equiv 0\pmod{13}$

Is there an easy way to find this?

And if I get an answer to this, am I okay to assume there are no other answers between $0$ and $13$?

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  • $\begingroup$ How about using the fact that $-10 \equiv 3 \pmod{13}$? $\endgroup$ – 2012ssohn Apr 29 '14 at 15:54
  • $\begingroup$ So 3x^7=10(mod13),3x^7=-3(mod13), x^7=-1(mod13), x=-1(mod13), x=12 $\endgroup$ – user146650 Apr 29 '14 at 16:00
  • $\begingroup$ Is that ok and also am I now all right to assume there are no other answers? $\endgroup$ – user146650 Apr 29 '14 at 16:03
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$3x^{10}\mod{13}=10x^3\mod{13}$

$-10x^{10}\mod{13}=10x^3\mod{13}$

$10(x^3+x^{10})\mod{13}=0\mod{13}$ This means:$\ $ $-x^3\mod{13}=x^{10}\mod{13}$

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  • $\begingroup$ Great! Really helpful, thank you. $\endgroup$ – user146650 Apr 29 '14 at 16:16
  • $\begingroup$ Why is it true that $-x^3\equiv x^{10}\pmod {13}$? Is there an obvious reason or was this left for the OP to further prove somehow? $\endgroup$ – user26486 Apr 29 '14 at 18:47
  • $\begingroup$ @mathh This was indeed something for the OP (original poster?) to solve... $\endgroup$ – gebruiker Apr 29 '14 at 19:16
  • $\begingroup$ @Aal Which way of proving it do you think is the best one? $\endgroup$ – user26486 Apr 29 '14 at 19:51
  • $\begingroup$ @Aal Could you tell me the way you've used to prove $-x^3\equiv x^{10}\pmod {13}$? I mean, I could split it into cases... When $13 \not\mid x$, we can prove that $x^7\equiv -1\pmod {13}$. Otherwise there's nothing to prove. So I could check the remainders $x^7$ can get when divided by $13$, but is there a more elegant way? This one is a bit tedious imho. $\endgroup$ – user26486 Apr 30 '14 at 12:03

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