Is it true that for any $x>2$ and $k,j \in N$ with $k>j$, $p_k(x)$ is larger than $p_j(x)$?

Question

Let $p_k(x)=(x+\sqrt{x^2-4})^k-(x-\sqrt{x^2-4})^k$.

Is it true that for any $x>2$ and $k,j \in N$ $k>j$, $p_k(x)$ is larger than $p_j(x)$?

Answer 1

Yes, it is true.

We can have $$\begin{align}\frac{d}{dk}&p_k(x)\gt 0\\&\iff \left(x+\sqrt{x^2-4}\right)^k\ln \left(x+\sqrt{x^2-4}\right)-\left(x-\sqrt{x^2-4}\right)^k\ln\left(x-\sqrt{x^2-4}\right)\gt 0\\&\iff \left(\frac{x+\sqrt{x^2-4}}{x-\sqrt{x^2-4}}\right)^k\gt\frac{\ln\left(x-\sqrt{x^2-4}\right)}{\ln\left(x+\sqrt{x^2-4}\right)}\\&\iff k\gt\log_X{\frac{\ln\left(x-\sqrt{x^2-4}\right)}{\ln\left(x+\sqrt{x^2-4}\right)}}\tag1\end{align}$$ where $$X=\frac{x+\sqrt{x^2-4}}{x-\sqrt{x^2-4}}\gt 1$$ Note here that $$\log_X{\frac{\ln\left(x-\sqrt{x^2-4}\right)}{\ln\left(x+\sqrt{x^2-4}\right)}}\lt 1\tag2$$ because $$\begin{align}&\log_X{\frac{\ln\left(x-\sqrt{x^2-4}\right)}{\ln\left(x+\sqrt{x^2-4}\right)}}\lt 1=\log_XX\\&\iff \frac{\ln\left(x-\sqrt{x^2-4}\right)}{\ln\left(x+\sqrt{x^2-4}\right)}\lt \frac{x+\sqrt{x^2-4}}{x-\sqrt{x^2-4}}\\&\iff \left(x+\sqrt{x^2-4}\right)\ln\left(x+\sqrt{x^2-4}\right)\gt \left(x-\sqrt{x^2-4}\right)\ln\left(x-\sqrt{x^2-4}\right)\end{align}$$ which is true.

It follows from $(1)(2)$ that $p_k(x)$ is increasing for $k\ge 1$.

Answer 2

Just another way. Note that the $p_k$ must be a solution to the recurrence with characteristic equation: $$\left((z-(x+\sqrt{x^2-4})\right)\left((z-(x-\sqrt{x^2-4})\right)=0$$ $\implies p_{k+2} = 2x\,p_{k+1}-4\,p_k$ which gives: $$\frac{p_{k+2}}{p_{k+1}} = 2x-4\frac{p_k}{p_{k+1}} > 4\left(1-\frac{p_k}{p_{k+1}}\right)$$ Now this inductively shows $\forall k \in \mathbb N, \quad \dfrac{p_{k+1}}{p_k}> 2$, with the base case $p_2/p_1 = 2x> 4>2$.