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First note that $\frac{1+i\sqrt{2}}{1-i\sqrt{2}}\in\mathbb{Q}(\sqrt{-2})$ is not a root of unity, because it does not equal $\pm 1$. Therefore Baker's famous theorem shows that, for some effectively computable constant $c>0$, $$\left|\left(\frac{1+i\sqrt{2}}{1-i\sqrt{2}}\right)^m-1\right|>m^{-c},\qquad m\geq 2.$$ (The variables $m$ and $n$ are integers in this post.) Applying this for $m=2n$, we get $$\left|\left(\frac{1+i\sqrt{2}}{1-i\sqrt{2}}\right)^n-1\right|\cdot\left|\left(\frac{1+i\sqrt{2}}{1-i\sqrt{2}}\right)^n+1\right|>(2n)^{-c},\qquad n\geq 1.$$ The first factor is less than $2$, hence multiplying both sides by the absolute value of $(1-i\sqrt{2})^n$, we infer $$\Bigl|(1+i\sqrt{2})^n+(1-i\sqrt{2})^n\Bigr|>\frac{1}{2}\cdot\frac{3^{n/2}}{(2n)^c},\qquad n\geq 1.$$ This means that the sequence $(a_n)$ grows exponentially: $$ |a_n|>\frac{1}{4}\cdot\frac{3^{n/2}}{(2n)^c},\qquad n\geq 1.$$ In particular, there are only finitely many $n$'s with $|a_n|\leq 100$, and these can be effectively bounded. Then, up to that bound, one can check with a computer (at least in theory) which $n$'s satisfy $|a_n|\leq 100$.