Given $x^5-x^3+x-2=0$, find $\lfloor x^6\rfloor$.

Question

If $\alpha$ is a real root of the equation $x^5-x^3+x-2=0$, find the value of $\lfloor\alpha^6\rfloor$, where $\lfloor x\rfloor$ is the least positive integer not exceeding $x$.

My approach is to bound the value of $\alpha^6=\alpha^4-\alpha^2+2\alpha$.

First I proved the function $f(x)=x^5-x^3+x-2$ is monotone increasing by derivative. Then I argue that $1<\alpha<\frac32$ as $f(1)<0$ and $f(\frac32)>0$.

Then I tried to create an upper and lower bound for $\alpha$, as such $$\alpha^6=\alpha^4-\alpha^2+2\alpha<\frac94\alpha^2-\alpha^2+2\alpha=\frac54\alpha^2+2\alpha<\frac{45}{16}+3=\frac{93}{16}$$ and $$\alpha^6=\alpha^4-\alpha^2+2\alpha>\alpha^3-\alpha^2+2\alpha=\alpha^2(\alpha-1)+2\alpha>2$$

Now we know that $\lfloor\alpha\rfloor\in\{2,3,4,5\}$. But I cannot proceed any further.

Any idea, help, hint, or answer is appreciated. Thank you.

Answer 1

Since $f'(x)=5x^4-3x^2+1$, which is always greater than $0$, $f$ is strictly increasing. Since, furthermore, $\lim_{x\to\pm\infty}f(x)=\pm\infty$, $f$ has one and only one real root. On the other hand, $f(1)=-1<0<24=f(2)$. So, the only real root $\alpha$ belongs to $(1,2)$ and $\lfloor\alpha\rfloor=1$.

Answer 2

$$x^5-x^3+x-2=x^5+x^2-(x^2-x+1)-(x^3+1)=$$ $$=(x^2-x+1)(x^3+x^2-x-2).$$ Now, we see that our equation has an unique real root $1<\alpha<2$ and from here $$[\alpha]=1.$$ Now, about your new problem.

Easy to see that for our root $\alpha$ we have $$1.205<\alpha<1.206,$$ which gives $$3<\alpha^6<4$$ and from here: $$[\alpha^6]=3.$$