5
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We have, $$\begin{aligned} (n^3-2n)^2 + (2n^2-1)^2 &= n^6 + 1\\ (n^5-2n^3+2n)^2 + (2n^4-2n^2+1)^2 &= n^{10} + 1\\ (n^7-2n^5+2n^3-2n)^2 + (2n^6-2n^4+2n^2-1)^2 &= n^{14} + 1\end{aligned}$$ and so on.

Q: What is a clever way to show that this pattern does in fact go on?

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here is one way

$n^{4k+2}+1 = (n^{2k+1}-i^{2k+1})(n^{2k+1}-(-i)^{2k+1}) \\ = (n-i)(n^{2k}+ \cdots + i^{2k})(n+i)(n^{2k}+\cdots + (-i)^{2k+1}))\\ = (n+i)(n^{2k}+ \cdots + i^{2k})(n-i)(n^{2k}+\cdots + (-i)^{2k+1}))\\ = (n^{2k+1}+2in^{2k}+\cdots +2i^{2k}n+i^{2k+1})(n^{2k+1}+2(-i)n^{2k}+\cdots +2(-i)^{2k}n+(-i)^{2k+1}) \\ = (n^{2k+1}-2n^{2k-1}+\cdots+2(-1)^kn)^2 + (2n^{2k}- \dots + (-1)^{k-1}2n^2+(-1)^k)^2$

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  • $\begingroup$ I should have known to factor it over the imaginary unit $i$. $\endgroup$ – Tito Piezas III Feb 18 '18 at 2:35

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