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# How many factors $$2$$ the number $$3^{x} + 1$$ has given that $$x$$ is an even number?

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 $$2$$.

Note by Lucas Tell Marchi
2 years, 7 months ago

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Great!

There is a one line solution to this problem, with a similar approach to what you did. Can you figure that out? Staff · 2 years, 7 months ago

Haven't found it yet :P · 2 years, 7 months ago

Hint: $$9 = 1 + 8$$ Staff · 2 years, 7 months ago

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

for some natural $$a$$ · 2 years, 7 months ago

Perfecto! Great job!

In fact, it is $$2 ( 1 + 4b )$$. Staff · 2 years, 7 months ago