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