Solve $99x^2 \equiv 1 \mod 125$ Solve $$x^{98} \equiv 99 \mod 125$$
Is there any easy way to solve equations like that? My observation is that from Euler's theorem we know that
$$ x^{100} \equiv 1 \mod 125 $$
so
$$x^{98} \equiv 99 \mod 125 \\ 
x^{100} \equiv 99x^2 \mod 125 \\
99x^2 \equiv 1 \mod 125$$
but what is general method how to deal with equations like that?

Start mod $5$, and then lift...
$$99 x^2 \equiv 4 x^2 \equiv (2x)^2 \equiv 1 \mod 5$$ 
so $2 x \equiv \pm 1 \mod 5$, i.e. $x \equiv 2$ or $3 \mod 5$.
If $x \equiv 2 \mod 5$, $x \equiv  2 + 5 y \mod 25$, and then
$$ 99 x^2 - 1 \equiv 5 y + 20 \equiv 0 \mod 25$$
$$ y + 4 \equiv 0 \mod 5$$
$$ y \equiv 1 \mod 5$$
So now $x \equiv 2 + 5 + 25 z \equiv 7 + 25 z \mod 125$, and then 
$$ 99 x^2 - 1 \equiv 25 z + 100 \equiv 0 \mod 125$$
$$ z + 4 \equiv 0 \mod 5$$
$$ z \equiv 1 \mod 5$$
Thus one solution is $x \equiv 2 + 5 + 25 \equiv 32 \mod 125$
