Prove that $\frac{2^{4n + 2} + 1}{5}$ is composed for all $n \ge 2$ The problem is to prove that $$\frac{2^{4n + 2} +1 }{5}$$ is composed for every $n \ge 2$.
I've tried this way: write $2^{4n + 2} + 1$ as $1 - (-4)^{2n + 1}$ and $5 = 1 - (-4)$, then $$\frac{2^{4n + 2} + 1}{5} = \frac{1 - (-4)^{2n + 1}}{1 - (-4)} = 1 - 4 + 16 - 64 + \cdots + (-4)^{2n}$$
How to proceed from this?
Thanks!

Check you can understand/prove the following
$$\frac{2^{4n+2}+1}5=\frac{3\cdot\left(2^{4n+1}-2^{4n}+\ldots-2^2+2\right)}5$$
But $\;5\nmid 6\;$ , so $\;5\;$ must divide the other factor as the whole thing is an integer since
$$4\cdot16^n+1=0\pmod 5$$
and thus the expression equals $\;6k\;,\;\;k\in\Bbb N\;$ ...composite.
