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!

One approach. First observe that we must have
$$2^{4n + 2} = - 1 \pmod 5$$
This is true because it is equivalent to 
$$4 * 16^n = -1 \pmod 5$$
And there are various ways of showing this holds for all $n$.
So we know the expression is an integer. As for compositeness, a hint is as follows. Note that 
$$4x^4 + 1 = (2x^2 + 2x + 1)(2x^2 - 2x + 1)$$
What does this say about the numerator?
