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

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

• Why can't you write it as $3^x + 1 = 2 \frac{3^x+1}{2}$? The fraction will simplify to an integer for all $x \geq 0$ Jun 18 at 5:54
• $3^x-1+2=2(1+3+\cdots +3^{x-1})+2=2(2+3+\cdots +3^{x-1})$ Jun 18 at 5:54
• The result is trivial by modular arithmetic : $3^x +1 \equiv 1^x +1 \equiv 0\,(mod\,2)$ , in case of unfamiliarity the binomial theorem is another straightforward way, as done in an answer below Jun 18 at 6:36

Hint : If $$x$$ is a natural number then $$3^x = (1+2)^x = 1 + x \cdot 2 + \cdots + 2^x$$ using binomial theorem.
• I see then we can factor two out of like so $$2(1 + x + 2 x \choose 2 + 4 x \choose 3 + \cdot + 2^{x-1})$$ but does that series simplify? Jun 18 at 6:34
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
If you're familiar with modular arithmetic, taking $$\pmod 2$$ gives $$1^x + 1 = 0 \pmod 2$$ as desired.