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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 \(\operatorname{\nu}_3(b_{2016})=3\).

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