Find $\lfloor {\alpha}^6 \rfloor$

Question

If $\alpha$ is a real root of the equation
$$x^5-x^3+x-2=0$$
find the value of $\lfloor {\alpha}^6 \rfloor$.

This one totally stumped me. We are asked to calculate $\lfloor {\alpha}^6 \rfloor$ without actually calculating the root or using wolfram alpha or any other calculator. I found that the above equation has only one real root by sketching its graph using calculus. I was also able to use the intermediate value theorem to conclude that $1<\alpha<2$ , but this is of little use while calculating $\lfloor {\alpha}^6 \rfloor$. Please help!

Answer 1

You might start by multiplying the equation by $x$ (or $\alpha$) to get $$\alpha^6=\alpha^4-\alpha^2+2\alpha$$ Next, dividing by $\alpha^2$ gives $$\alpha^4=\alpha^2-1+\frac2\alpha$$ and combining these two facts we have $$ \alpha^6 = 2\alpha - 1 + \frac 2\alpha $$ whose range on $(1,2)$ is small enough to be dealt with by the floor.

Answer 2

One important thing is that $$ \begin{aligned} f^{\prime}(x) & =5 x^4-3 x^2+1 \\ & =5\left(x^2-\frac{3}{10}\right)^2+\frac{5}{100} \\ & >0 \end{aligned} $$ implies that $f(x)$ is strictly increasing. Noting that $f(1)=-1$ and $f(\sqrt2)=3\sqrt2-1$, we know that $f(x)=0$ has only a unique root $\alpha $ satisfying $1<\alpha <\sqrt2$. Reciprocal reverses the inequality that $\frac{1}{\sqrt{2}}<\frac{1}{\alpha}<1$. Hence $$\alpha+ \frac{1}{\alpha} <\sqrt2+1 \cdots (1) $$

$$\begin{aligned} 2&=\alpha^5-\alpha^3+\alpha =\frac{\alpha\left[1-\left(-\alpha^2\right)^3\right]}{1-\left(-\alpha^2\right)}=\frac{\alpha\left(1+\alpha^6\right)}{1+\alpha^2} & \\\Rightarrow \quad & 1+\alpha^6=2\left(\alpha+\frac{1}{\alpha}\right)<2(\sqrt{2}+1)\end{aligned}$$

On the other hand, $$\quad 1+\alpha^6=2\left(\alpha+\frac{1}{\alpha}\right)\le 2 \sqrt{\alpha \cdot \frac{1}{\alpha}}=4 \cdots (2) $$

Combining them yields $$ \begin{aligned} 4 & \leqslant 1+\alpha^6<2(\sqrt{2}+1) \\ \Rightarrow \quad & 3 \leqslant \alpha^6<2 \sqrt{2}+1<4 \\ \Rightarrow \quad & \left\lfloor\alpha^6\right\rfloor =3\end{aligned} $$