# Find all answers to $3x^{10}=10x^3\pmod{13}$

Number theory related question.

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

$0$ 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 $13$?

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

$3x^{10}\mod{13}=10x^3\mod{13}$
$-10x^{10}\mod{13}=10x^3\mod{13}$
$10(x^3+x^{10})\mod{13}=0\mod{13}$ This means:$\$ $-x^3\mod{13}=x^{10}\mod{13}$
• 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? – user26486 Apr 29 '14 at 18:47
• @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. – user26486 Apr 30 '14 at 12:03