First of all, given that $x$ is a whole even number, let's write $x = 2m$ for some natural $m$. Then observe that we are allowed, according to Newton, to write

$3^{2m} = (1+2)^{2m} = \sum_{n=0}^{2m} \begin{pmatrix} 2m\\ n \end{pmatrix} 2^{n} = 1 + \sum_{n=1}^{2m} \begin{pmatrix} 2m\\ n \end{pmatrix} 2^{n}$

And notice that

$\sum_{n=1}^{2m} \begin{pmatrix} 2m\\ n \end{pmatrix} 2^{n}$

is an even number, since the factor $2$ happens to appear in all of the terms (and this is only possible because $0 < n < 2m + 1$).

Now, this is $3^{x}$. Adding the $1$ we get

$3^{2m} + 1 = 2 + \sum_{n=1}^{2m} \begin{pmatrix} 2m\\ n \end{pmatrix} 2^{n} = 2 \times \left [ 1 + \sum_{n=1}^{2m} \begin{pmatrix} 2m\\ n \end{pmatrix} 2^{n-1} \right ]$

But

$1 + \sum_{n=1}^{2m} \begin{pmatrix} 2m\\ n \end{pmatrix} 2^{n-1}$

is an odd number, since

$\begin{pmatrix} 2m\\ 1 \end{pmatrix}$

is even (for $n = 1$, where $2^{n-1} = 1$) and for all the other terms $2^{n-1}$ is even. Therefore

$1 + \sum_{n=1}^{2m} \begin{pmatrix} 2m\\ n \end{pmatrix} 2^{n-1}$

has no factors $2$ and

$2 \times \left [ 1 + \sum_{n=1}^{2m} \begin{pmatrix} 2m\\ n \end{pmatrix} 2^{n-1} \right ]$

has all the factors $2$ in evidence. Then $3^{x} + 1$ for an even $x$ has only one factor $2$.

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## Comments

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TopNewestGreat!

There is a one line solution to this problem, with a similar approach to what you did. Can you figure that out?

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Haven't found it yet :P

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Hint:$9 = 1 + 8$Log in to reply

$(1+8)^{x} = 4a + 1 \;\;\; \Rightarrow \;\;\; (1+8)^{x} + 1 = 4a + 2 = 2(2a+1)$

for some natural $a$

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In fact, it is $2 ( 1 + 4b )$.

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