<s>[INST] How to reduce congruence power modulo prime? If I have a congruence equation, says 
$$x^{15} - x^{10} + 4x - 3 \equiv 0 \pmod{7}$$
Then can I use Fermat's little theorem like this:
$$(x^{6})^2 \cdot x^3 - x^6 \cdot x^4 + 4x - 3 \equiv 0 \pmod{7}$$
$$ x^3 - x^4 + 4x - 3 \equiv 0 \pmod{7}$$
Update
Should it be 
$$x^{14}x - x^7x^3 - 4x - 3 \equiv 0 \pmod{7}$$
$$x^2x - x.x^3 - 4x - 3 \equiv 0 \pmod{7}$$
$$x^3 - x^4 - 4x - 3 \equiv 0 \pmod{7}$$
?
Thanks, 
 [/INST] Yes, if you're looking for solutions of the equation mod $7$ then, since $\rm\:x=0\:$ is not a solution, you can in fact deduce that $\rm\:x^6 = 1\:$. If you couldn't exclude $\rm\:x=0\:$ then you'd instead need $\rm\:x^7 = x\:.$
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