Since $$1+i = \sqrt2 (\cos \dfrac{\pi}{4} + i \sin \dfrac{\pi}{4})$$ and $$1-i = \sqrt2 (\cos \dfrac{\pi}{4} - i \sin \dfrac{\pi}{4})$$ then $$\begin{split} (1+i)^n + (1-i)^n &= \left( \sqrt2 (\cos \dfrac{\pi}{4} + i \sin \dfrac{\pi}{4}) \right)^n + \left( \sqrt2 (\cos \dfrac{\pi}{4} - i \sin \dfrac{\pi}{4}) \right)^n \\ &= 2^{\frac{n}{2}} ( \cos \dfrac{n \pi}{4} + i \sin \dfrac{n \pi}{4} ) + 2^{\frac{n}{2}} ( \cos \dfrac{n \pi}{4} - i \sin \dfrac{n \pi}{4} ) \\ &= 2^{\frac{n}{2}+1} \cos \dfrac{n \pi}{4} \end{split}$$ For \(\cos \dfrac{n \pi}{4}\) is positive, $n = 8k, \; n = 8k+1$ or $n=8k+7$, where $k$ is a non-negative integer.
Hence the smallest positive integer $n$ is $$n = 71$$
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