Check if any number of this form is composite

Question

I got this problem to solve saying : Given that $n>2$ and $n$ is a natural number, prove that $y = 2^n + (-1)^{n+1}$ is a composite number. I took the above and I did the following: $$y = 2^n + (-1)^n(-1)^1 = 2^n - (-1)^n$$ and because of the factorization form of, lets say: $$a^n - b^n = (a-b)(a^{n-1} + ba^{n-2} +\cdots+ b^{n-2}a +b^{n-1})$$I get: $$y=3(2^{n-1}-2^{n-2} + ... +(-1)^{n-2}2 + (-1)^{n-1}).$$

If the result was somewhat like $y=3\cdot5$ , I would for a fact know that $y$ is a composite (not a prime number). Do I have to say/prove anything else for the second multiplier? Thank you for your help

Answer 1

You need say nothing about the second factor. You have shown $3$ divides $y$ and this is sufficient to show that $y$ is composite.

Answer 2

Yes, to prove that $\,3(y/3)$ is composite you need to prove that $\,y/3\,$ is not $\,\pm1$ or $\,0.\,$ This follows since $\,n\ge 3\,$ implies that $\,2^n\ge 8\,$ so $\,y = 2^n + (-1)^{n+1} \ge 8\pm 1 \ge 7,\,$ so $\,y/3 \ge 2.$