# Find sufficient and necessary conditions in which $f(x)$ is a natural number

Let us consider the function $$f(x)=(-√3+2)^{2^{x-2}}+(√3+2)^{2^{x-2}}$$ where $x≥3$.

We know for example that if $x$ is an integer, then $f(x)$ is also an integer.

My question is: Find sufficient and necessary conditions in which $f(x)$ is a natural number.

1st claim :

Let $k \in \mathbb{N}$ with $k \geq 2$. Then :

$$\Big( 2\cosh(x) = k \Big) \, \Leftrightarrow \, x = \ln \Big( \frac{k+\sqrt{k^2-4}}{2} \Big).$$

Let $k \in \mathbb{N}$ with $k \geq 2$. We have :

\begin{align*} 2\cosh(x) = k &\Leftrightarrow {} e^{2x} - ke^{x} + 1 = 0 \\ \end{align*}

Let $u=e^{x} > 0$. We solve $u^2 - k u + 1 =0$ and find that $u=e^{x}=k+\sqrt{k^2-4}$.

2nd claim :

For all $x \in \mathbb{R}$ with $x \geq 3$,

$$f(x) = 2 \cosh \big( 2^{x-2} \ln(\sqrt{3}+2) \big).$$

Just note that : $\ln(-\sqrt{3}+2) = - \ln(\sqrt{3}+2)$.

3rd claim :

Let $k \in \mathbb{N}$ with $k \geq 2$. Then :

$$f(x) =k \; \Leftrightarrow \; x = \frac{1}{\ln(2)} \Bigg( (\ln \circ \ln) \Big( \frac{k+\sqrt{k^2-4}}{2} \Big) - (\ln \circ \ln)(\sqrt{3}+2) \Bigg) + 2.$$

We have :

\begin{align*} f(x) = k &\Leftrightarrow {} 2^{x-2}\ln(\sqrt{3}+2) = \underbrace{\ln\Big( \frac{k+\sqrt{k^2-4}}{2} \Big)}_{>0} \\[2mm] &\Leftrightarrow (x-2)\ln(2) + (\ln \circ \ln)(\sqrt{3}+2) = (\ln \circ \ln)\Big( \frac{k+\sqrt{k^2-4}}{2} \Big) \\[2mm] \end{align*}

Which gives the expected value for $x$.

Then, you can verify that if $x$ is given as in the third claim, then $f(x)$ is an integer.

• @ jibounet: Is this analysis also true for this function: $$h(x)=(((-√3+2)^{2^{x-2}}+(√3+2)^{2^{x-2}})/(2^{x}-1))$$ – DER Oct 29 '14 at 16:36
• @DER : No, it is not because for $x$ as in claim 3, $2^x -1$ will not be an integer. – jibounet Oct 29 '14 at 16:39
• @ jibounet: No I mean by using this method for this case: $h(x)=m$ to find another $x$ – DER Oct 29 '14 at 16:42
• @DER : No, I don't think the same method will apply to $h$. – jibounet Oct 29 '14 at 16:46
• @ jibounet: Are you sure about the second claim. – DER Oct 29 '14 at 18:00