Final Round of National Olympiad

Note that $3^{2n} - 3^{n+1} + 3^n - 3 = (3^n - 3)(3^n + 1)$.

If $n$ is odd then $3^n - 3 = 3(3^{n-1} - 1) = 3\left(3^{\frac{n-1}{2}} - 1\right)\left(3^{\frac{n-1}{2}} + 1\right)$.

As all powers of $3$ are odd, $3^{\frac{n-1}{2}} - 1$ and $3^{\frac{n-1}{2}} + 1$ are consecutive even numbers. One of these numbers must be divisible by $4$, whence their product is divisible by $8$. Thus $8 \mid 3^n - 3$, implying that $4 \mid 3^n + 1$. Consequently, $32 \mid 3^{2n} - 3^{n+1} + 3^n - 3$.

Assume now $n$ being even. Note that
$$3^{2n} - 3^{n+1} + 3^n + 1 = 3^{2n} - 3 \cdot 3^n + 3^n + 1 = (3^n)^2 - 2 \cdot 3^n + 1 = (3^n - 1)^2.$$
Similarly to the previous case, $3^n - 1 = (3^{\frac{n}{2}} - 1)(3^{\frac{n}{2}} + 1)$ where factors in the r.h.s. are consecutive even numbers. Hence $8 \mid 3^n - 1$, implying that $64 \mid 3^{2n} - 3^{n+1} + 3^n + 1$. Consequently, $32 \mid 3^{2n} - 3^{n+1} + 3^n + 1$.