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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, 10 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?

Calvin Lin Staff - 2 years, 10 months ago

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Haven't found it yet :P

Lucas Tell Marchi - 2 years, 10 months ago

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Hint: \( 9 = 1 + 8 \)

Calvin Lin Staff - 2 years, 10 months ago

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@Calvin Lin \[(1+8)^{x} = 4a + 1 \;\;\; \Rightarrow \;\;\; (1+8)^{x} + 1 = 4a + 2 = 2(2a+1) \]

for some natural \(a\)

Lucas Tell Marchi - 2 years, 10 months ago

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@Lucas Tell Marchi Perfecto! Great job!

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

Calvin Lin Staff - 2 years, 10 months ago

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