Prove that for all $x$ there exists an $n$ such that this expression is a power of $2$

Question

Would just like to preface this with, I do not know if this is solveable or not as its not homework its just something I was doing for fun while looking at the collatz conjecture so I don't expect anyone to solve it and wouldn't want anyone to waste their time on this question.

I am trying to show that

$$\forall_{x\in Z_+}{ \exists_{n \in Z} : 2(x+1)(\frac{3}{2})^n - 2 = 2^k}$$ Below is the same expression just in a different form.

$$ \frac{(x+1)3^n - 2^n}{2^{n-1}} = 2^k $$ In other words, for every positive odd integer value of x, I must show that there exists a value of n such that the expression is a power of 2.

My thoughts moving forward from here are to try and find an integer value of n in terms of x that allows me to write the expressions in terms of a power of 2 that is easier said than done however so my main question is, are there any "simple" techniques and/or known methods that I am missing so far in order to progress or have I reached the bit that makes it unsolved. Thank you.

Sorry if this was the wrong place to ask this.

Edit : x will always be odd.

Answer 1

This doesn't seem possible in general. If $x$ is even and $x+1$ is not a multiple of $3$, then the only possible choices for $n$ that make the left-hand side an integer are $n=1$ and $n=0$. In other words, in this case one of the two numbers $2x$ and $3x+1$ would need to be a power of $2$, and this won't always be the case. So $x=6, 10, 12, 14, 18, 20, 22, 24, 26, 28, 30,\dots$ are all counterexamples.

Answer 2

Rewriting your equation $$ 2(x+1)(\frac{3}{2})^n - 2 = 2^k \\ (x+1)3^n=2^n(2^{k-1}+1)$$ then $$ {x+1\over 2^n}={ 2^j+1\over3^n} \tag 1 $$ (using one letter $j$ for 3 letters $k-1$).

We compare now the lhs and rhs.

• On lhs: because denominators have no commnon factor we must have with some $t$ that $x$ has the form $ x=t \cdot 2^n-1 $ and then the equation looks like this: $$ t = {2^j+1 \over 3^n} \tag 2 $$
• Now, on the rhs, recalling Fermat, Euler and the "lifting the exponent lemma", we have the factoring of $2^j+1$ with respect to $3$: (I use $\{a,b\}$ for $\nu_b(a)$ and $[a:b]= 1$ if $a$ is divisible by $b$ and $0$ if not) : $$\{2^j+1,3\} = [j-1:2](1+ \{j,3\}) \tag 3 $$ So to have the factor $3^n$ we need that $j$ is odd and $j$ itself contains the factor $3^{n-1}$, so $j$ must be of the form $j=i3^{n-1}$ with any odd $i$.
  Solutions should be $$ t = {2^{i \cdot 3^{n-1}}+1 \over 3^n} \qquad i \text{ is odd}\\ \tag 4 $$

Basic integer solutions on the rhs are thus -with $i=1$ - : $$ 2^{1 \cdot 3^{n-1}}+1 = \small \begin{array} {rl} n & \text{factorization} \\ \hline 1 & 3 \\ 2 & 3^2 \\ 3 & 3^3.19 \\ 4 & 3^4.19.87211 \\ 5 & 3^5.19.163.87211.135433.(big) \end{array} \tag {5a} $$ But the exponent of $2$ in (eq 4) can have additionally any odd factor $i$, for instance $i=5$: $$ 2^{5 \cdot 3^{n-1}}+1 = \small \begin{array} {rl} n & \text{factorizing} \\ \hline 1 & 3.11 \\ 2 & 3^2.11.331 \\ 3 & 3^3.11.19.331.18837001 \\ 4 & 3^4.11.19.331.811.15121.87211.(big) \\ 5 & 3^5.11.19.163.331.811.6481.9721.15121.87211.(big) \end{array} \tag {5b} $$ and with $i=7$ $$ 2^{7 \cdot 3^{n-1}}+1 = \small \begin{array} {rl} n & \text{factorizing} \\ \hline 1 & 3.43 \\ 2 & 3^2.43.5419 \\ 3 & 3^3.19.43.5419.(big) \\ 4 & 3^4.19.43.379.5419.87211.(big) \\ 5 & 3^5.19.43.163.379.5419.87211.(big) \end{array} \tag {5c} $$

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Back to your original question:

Using the second example $j=5 \cdot 3^{n-1}$ we have for your $k$ and $x$: $$k=5 \cdot 3^{n-1}+1 \tag {6a}$$ and $$x=t \cdot 2^n-1 = {2^j+1\over 3^n} \cdot 2^n-1 \\ = {2^{5 \cdot 3^{n-1}}+1\over 3^n} \cdot 2^n-1 \tag {6b}$$

Finally, to evaluate your first formula using $x$ and $k$ we find, comparing the lhs with rhs, that they are equal and we show the exponent of $2$: $$ \small \begin{array} {} n & \{lhs,2\} & \{ rhs,2\} \\ \hline 1 & 16 & 16 \\ 2 & 46 & 46 \\ 3 & 136 & 136 \\ 4 & 406 & 406 \\ 5 & 1216 & 1216 \end{array} \tag {7} $$

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So the final answer to your question: there are solutions for your equation, but only for $x$ from a subset of $\mathbb N$ parametrized by $(n,i)$, namely: $$x= {2^{i \cdot 3^{n-1}}+1\over 3^n} \cdot 2^n-1 \qquad \left(i \in \mathbb N \backslash 2 , n \in \mathbb N\right) \tag 8$$

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The top-left of the matrix of solutions for $x$ depending on $n$ (column-index) and $i=2h-1$ ($h$ indicate rows) (row&column indexes begin at $1$) is $$x \in \small \begin{array} {r|rrrrr} h \backslash n& 1&2&3&\cdots \\ \hline 1&1 & 3 & 151& \cdots \\ 2&5 & 227 & 39768215 & \cdots\\ 3&21 & 14563 & 10424999137431 & \cdots\\ 4&85 & 932067 & 2732850973882896535 & \cdots\\ 5&341 & 59652323 & 716400485697558029455511& \cdots \\ \vdots &\vdots &\vdots & \ddots \end{array} \tag 9 $$ As far as my answer is correct at all, $x$ must be element of this sparse subset of $\mathbb N$.