Find the smallest non-zero natural number $n$ such that 2016 divides $20^{n}-16^{n}$.
We start by factoring 2016 into a product of prime factors: $2016=2^{5} \times 3^{2} \times 7$. We need to find the integers $n \geqslant 1$ such that the integer $20^{n}-16^{n}=2^{2 n}\left(5^{n}-4^{n}\right)$ is divisible by $2^{5}, 3^{2}$, and 7. This is exactly the set of integers such that $2 n \geqslant 5$, $5^{n} \equiv 4^{n}\left(\bmod 3^{2}\right)$, and $2^{n} \equiv 4^{n}(\bmod 7)$.

Since $4 \times 7 \equiv 1\left(\bmod 3^{2}\right)$ and $4 \times 2(\bmod 7)$, this is equivalent to $n \geqslant 3$, $(-1)^{n} \equiv (5 \times 7)^{n} \equiv (4 \times 7)^{n} \equiv 1^{n} \equiv 1\left(\bmod 3^{2}\right)$, and $3^{n} \equiv (5 \times 2)^{n} \equiv (4 \times 2)^{n} \equiv 1^{n} \equiv 1(\bmod 7)$.

Let $\omega$ be the order of -1 modulo $3^{2}$ and $\omega^{\prime}$ be the order of 3 modulo 7. The integer $n$ we are looking for is the smallest multiple of $\omega$ and $\omega^{\prime}$ such that $n \geqslant 3$. Now, $(-1)^{1} \not \equiv 1\left(\bmod 3^{2}\right)$ and $(-1)^{2} \equiv 1$ $\left(\bmod 3^{2}\right)$, so $\omega=2$.

Similarly, $3^{1} \not \equiv 1(\bmod 7)$, $3^{2} \equiv 2 \not \equiv 1(\bmod 7)$, and $3^{3} \equiv-1 \not \equiv 1(\bmod 7)$, so $\omega^{\prime}>3$, and the little Fermat theorem ensures that $\omega^{\prime}$ divides 6. This means that $\omega^{\prime}=6$. The smallest integer $n$ we are looking for is therefore $n=6$.