When $n$ is an odd natural composite number, $\frac{2^n + 1}{3}$ is always composite. Prove this conjecture.
I'm not 100% sure but I let $n$ = $2c+1$ = $ab$, where $c$ is a natural number including $0$, and $a$ and $b$ are natural numbers, where $1<a$ and $b<n$.
$\frac{2^{ab} + 1}{3}$ = $\frac{(2^b + 1)(1-2^b+2^{2b}+.....+2^{b(a-1)})}{3}$. Now let $x$ = $2^b + 1$, $y$ = $(1-2^b+2^{2b}+.....+2^{b(a-1)})$.
I know that I can prove that the numerator is composite as $1<b<n$, so $1$<$x$< $2^n + 1$, which implies that $1$<$y$<$2^n + 1$, which means that the numerator is composite.
I am stuck with the divided by $3$ part. I guess I should use the fact that $n$ is odd but I am not sure how valid my proof is. Is it do with the fact that the numerator is somehow a multiple of 3 when $n$ is odd, which I guess I could prove using induction ?
How bad is my proof ? :( Thanks guys!