Problem. Use Tutte's 1-factor theorem to prove that every 3-regular graph with no bridges has a 1-factor.

I honestly have no idea how to approach this problem. I have drawn a couple of graphs to gain some insight to no avail. Any hints or tips will be greatly appreciated.

Wikipage on modular arithmetic is not bad.

When $b$ is huge, and $a$ and $c$ are coprime, Euler's theorem applies:
$$
    a^b \equiv a^{b \, \bmod  \, \phi(c)} \, \bmod c
$$
For the example at hand, $\phi(21) = \phi(3) \times \phi(7) = 2 \times 6 = 12$.
$$
\Rightarrow 844325 \bmod 12 = 5,\ \text{so}\ 5^5 = 5 \times 25^2 \equiv 5 \times 4^2 = 80 \equiv 17 \mod 21
$$.

When $a$ and $c$ are coprime, but $0<b<\phi(c)$, repeated squaring (or using other compositions of powers) is the fastest way to go (manually):
$$
\begin{eqnarray}
   5^4 \equiv 5 \times 5^3 \equiv 5 \times 24 \equiv 19 &\pmod{101}\\
   19^4 \equiv (19^2)^2 \equiv 58^2 \equiv (-43)^2 \equiv 1849 \equiv 31 &\pmod{101} \\
   31^4 \equiv (31^2)^2 \equiv (961)^2 \equiv 52^2 \equiv 2704 \equiv 78 &\pmod{101} \\
   5^{69} \equiv 5 \times 5^4 \times ((5^4)^4)^4 \equiv 5 \times 19 \times 78 \equiv 5 \times 19 \times (-23)\\
 \equiv 19 \times (-14) \equiv -266 \equiv  37 & \pmod{101} 
\end{eqnarray}
$$

When $a$ and $c$ are not coprime, let $g = \gcd(a,c)$. Let $a = g \times d$ and $c = g \times f$, then, assuming $b > 1$:
$$
   a^b \bmod c = g^b \times d^b \bmod (g \times f) = ( g \times (g^{b-1} d^b \bmod f) ) \bmod c
$$
In the example given, $\gcd(6,14) = 2$. So $2^{102} \times 3^{103} \mod 7$, using Euler'r theorem, with $\phi(7) = 6$,  and $102 \equiv 0 \mod 6$, $2^{102} \times 3^{103} \equiv 3 \mod 7$, so $6^{103} \equiv (2 \times 3) \equiv 6 \mod 14 $.


