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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

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    $\begingroup$ 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 $x \geq 0$ $\endgroup$
    – bigbang
    Jun 18 at 5:54
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    $\begingroup$ $3^x-1+2=2(1+3+\cdots +3^{x-1})+2=2(2+3+\cdots +3^{x-1})$ $\endgroup$
    – Asher2211
    Jun 18 at 5:54
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    $\begingroup$ 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 $\endgroup$
    – Amartya S
    Jun 18 at 6:36
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Hint : If $x$ is a natural number then $3^x = (1+2)^x = 1 + x \cdot 2 + \cdots + 2^x$ using binomial theorem.

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  • $\begingroup$ 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? $\endgroup$ Jun 18 at 6:34
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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

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

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