Highest Powers of 3 and $\left(1+\sqrt{2}\right)^n$

Let $\left(1+\sqrt{2}\right)^{2012}=a+b\sqrt{2}$, where $a$ and $b$ are integers. What is the greatest common divisor of $b$ and $81$?
Source: 2011-2012 SDML High School 2a, problem 15.

Let $(1+\sqrt{2})^n = a_n + b_n \sqrt{2}$. I've thought about this some more, and there's a nice way to describe the highest power of $3$ that divides $b_n$. This is probably outside of the scope of the intended solution, however.

First note that $(1-\sqrt{2})^n = a_n - b_n \sqrt{2}$, and so from $(1+\sqrt{2})(1-\sqrt{2})=-1$ we get $(1+\sqrt{2})^n (1-\sqrt{2})^n = {(-1)}^n$. This gives $${a_n}^2 - 2 {b_n}^2 = {(-1)}^n.$$ Now define the highest power of a prime $p$ that divides $n$ to be $\operatorname{\nu}_p(n)$.
From cubing and using the above result it's straightforward to prove that if $\operatorname{\nu}_3(b_n) = k > 0$ then $\operatorname{\nu}_3(b_{3n}) = k+1$.
Note $(1+\sqrt{2})^4 = 17 + 12\sqrt{2} \equiv -1+3\sqrt{2} \pmod{3^2}$. Cubing and using the first formula as before, we can in fact show that $$(1+\sqrt{2})^{4\cdot 3^n} \equiv -1 + 3^{n+1}\sqrt{2} \pmod{3^{n+2}},$$ and squaring we also have $$(1+\sqrt{2})^{8\cdot 3^n} \equiv 1 + 3^{n+1}\sqrt{2} \pmod{3^{n+2}}.$$ Now assume $\operatorname{\nu}_3(b_m) = k, \operatorname{\nu}_3(b_n) = l$ and $k\neq l$. From the top formula if $3 | b_i$ then $3 \not{|} a_i$, and it follows that $$\operatorname{\nu}_3(b_{m+n}) = \min(k,l).$$ Putting this all together, write $n = 4\cdot m +k$, where $0\leq k <4$. If $k\neq 0$, then $\operatorname{\nu}_3(b_{n}) = 0$. If $k=0$, let the base-3 expansion of $m$ be $a_i \cdot 3^i + \ldots + a_0$. Then $$\operatorname{\nu}_3(b_{n}) = \min_{a_j \neq 0} j+1 .$$
For $n=2012$, we have $2012 = 4\cdot 503 = 4\cdot(2\cdot 3^5 + 3^2 + 2\cdot 3 + 2)$ and so $\operatorname{\nu}_3(b_{2012})=1$. We don't actually need to compute the entire base-3 expansion for 503, of course; we only need to observe that it's not divisible by 3.

For $n=2016$, we have $2016 = 4\cdot 504 = 4\cdot(2\cdot 3^5 + 2\cdot 3^2)$ and so $\operatorname{\nu}_3(b_{2016})=3$.