On near-Pythagorean triples $(n^5-2n^3+2n)^2 + (2n^4-2n^2+1)^2 = n^{10} + 1$ 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?

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