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 $13$?

  • $\begingroup$ 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



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

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.