Factoring $2$ from $3^x + 1$

Question

I was thinking about how I could prove algebraically that $3^x + 1$ has a factor of $2$ by expressing it in the form $2y$, where $y$ can be solved in terms of $x$ in simplest terms.

By basic intuition we know that it is factorable because $3^x$ is definitely odd, so adding 1 should make it even and thus divisible by $2$. However, I have not been able to simplify it. How would one go about this? Thanks

Answer 1

Hint : If $x$ is a natural number then $3^x = (1+2)^x = 1 + x \cdot 2 + \cdots + 2^x$ using binomial theorem.

Answer 2

Assuming unfamiliarity of the OP with modular arithmetic I use an inductive approach :

Assuming $2 \mid (3^k + 1)$ for some $k \in \mathbb{Z}^{+}$, we've $3^{k+1}+1=3(3^k +1)-2 =3(2\xi)-2=2(3\xi-1) \implies 2 \mid (3^{k+1} +1)$

$\xi$ is an arbitrary positive integer

Answer 3

If you're familiar with modular arithmetic, taking $\pmod 2$ gives $1^x + 1 = 0 \pmod 2$ as desired.