Q:

Congruence equation, I have no idea how to do it.

I have here this equation:
$10x^{84} + 3x + 7 ≅ 0 (mod 35) $
I can't really do $y=x^{84}$ since I have a simple $x$ too, any other ideas? Thanks! :)

A:

First notice that:
$$10x^{84}+3x+7\equiv 0 \mod 35$$
implies $5$ divides $3x+7$ and then $x=1 \mod 5$ 
But $7$ must divide $3x(x^{83}+1)$ hence $x\equiv 0 \mod 7$ or $x^{83}\equiv -1 \mod 7$ using Fermat's little theorem $x^{84}\equiv (x^{14})^6\equiv 1\mod 7$ hence $x^{83}\equiv x^{-1}$ we conclude that $x\equiv 0 \mod 7$ or $x\equiv -1 \mod 7$
Finally $x\equiv 21 \mod 35$ or $x\equiv 6 \mod 35$.

