Prove that $(4^p+1)/5$ is a composite integer for $p$ a prime greater than 5.

Question

So far I have found that $(4^p+1)/5=\sum_{n=0}^{p-1} (-1)^n4^n$, not sure where to go from here. Was thinking about trying to find some base for which $(4^p+1)/5$ is not a pseudoprime but have had no luck.

Answer 1

As $p$ is odd:

$$ 4 x^4 + y^4 = (2x^2 + 2 xy + y^2)(2 x^2 - 2xy + y^2) $$

Taking $$ t = \frac{p-1}{2} $$ we have $$ x = 2^t, $$ $$ x^2 = 4^t $$ $$ x^4 = 4^{2t} = 4^{p-1}, $$ $$ 4 x^4 = 4^p. $$ Then take $$ y = 1. $$

Answer 2

Set $x=\frac{1}{2}(p-1)\ge 2$, then $$ 1+4^p=(1+4\cdot 2^{4x}+2^{2x+2})-2^{2x+2}=(1+2^{2x+1}+2^{x+1})(1+2^{2x+1}-2^{x+1}). $$