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