How to solve the congruence $y^{31}\equiv 3 \mod{100}$ $\phi (100) = 40$
Hence:
$y^{31}\equiv y^{-9} \equiv 3\mod{100}$
$y^{9}\equiv 67 \mod{100}$
However I do not know where to go from here.

This is a kind of tedious solution:
$$x^{31}\equiv 3\pmod{100}\iff \begin{cases}x^{31}\equiv 3\pmod{25}\\ x^{31}\equiv 3\pmod{4}\end{cases}$$
$$\iff \begin{cases}x^{31}\equiv 3\pmod{25}\\ x\equiv 3\pmod{4}\end{cases}$$
Clearly $\gcd(x,25)=1$, so by Euler's theorem:
$$\iff \begin{cases}x^{11}\equiv 3\pmod{25}\\ x\equiv 3\pmod{4}\end{cases}$$
$2$ is a primitive root mod $5^2$, because $\gcd(2,25)=1$ and $\phi(25)=20=2^2\cdot 5$ and 
$2^{\frac{20}{2}}\equiv 2^{10}\equiv 1024\equiv -1\not\equiv 1\pmod{25}$
and $2^{\frac{20}{5}}\equiv 2^{4}\equiv 16\not\equiv 1\pmod{25}$.
Let $x\equiv 2^k\pmod{25}$ for some $k\in\mathbb Z^+$.
$$2^{11k}\equiv 3\equiv 2^7\pmod{25}\iff 11k\equiv 7\equiv -33\pmod{20}$$
$$\stackrel{:11}\iff k\equiv -3\equiv 17\pmod{20}$$
Therefore $x\equiv 2^{17}\equiv 2^{10}2^{7}\equiv (1024)(128)\equiv (-1)(3)\equiv -3\equiv 22\pmod{25}$,
so by Chinese Remainder Theorem $x\equiv 47\pmod{100}$.
