, 2004

Solution:

Suppose $\alpha$ is a real root of the given equation. Then
$$\alpha^{5}-\alpha^{3}+\alpha-2=0$$
This gives $\alpha^{5}-\alpha^{3}+\alpha-1=1$ and hence $(\alpha-1)\left(\alpha^{4}+\alpha^{3}+1\right)=1$.

Observe that $\alpha^{4}+\alpha^{3}+1 \geq 2 \alpha^{2}+\alpha^{3}=\alpha^{2}(\alpha+2)$. If $-1 \leq \alpha<0$, then $\alpha+2>0$, giving $\alpha^{2}(\alpha+2)>0$ and hence $(\alpha-1)\left(\alpha^{4}+\alpha^{3}+1\right)<0$. If $\alpha<-1$, then $\alpha^{4}+\alpha^{3}=\alpha^{3}(\alpha+1)>0$ and hence $\alpha^{4}+\alpha^{3}+1>0$. This again gives $(\alpha-1)\left(\alpha^{4}+\alpha^{3}+1\right)<0$.

The above reasoning shows that for $\alpha<0$, we have $\alpha^{5}-\alpha^{3}+\alpha-1<0$ and hence cannot be equal to $1$. We conclude that a real root $\alpha$ of $x^{5}-x^{3}+x-2=0$ is positive (obviously $\alpha \neq 0$).

Now using $\alpha^{5}-\alpha^{3}+\alpha-2=0$, we get
$$\alpha^{6}=\alpha^{4}-\alpha^{2}+2 \alpha$$
The statement $\left[\alpha^{6}\right]=3$ is equivalent to $3 \leq \alpha^{6}<4$.

Consider $\alpha^{4}-\alpha^{2}+2 \alpha<4$. Since $\alpha>0$, this is equivalent to $\alpha^{5}-\alpha^{3}+2 \alpha^{2}<4 \alpha$. Using the relation (1), we can write $2 \alpha^{2}-\alpha+2<4 \alpha$ or $2 \alpha^{2}-5 \alpha+2<0$. Treating this as a quadratic, we get this is equivalent to $\frac{1}{2}<\alpha<2$.

Now observe that if $\alpha \geq 2$ then $1=(\alpha-1)\left(\alpha^{4}+\alpha^{3}+1\right) \geq 25$ which is impossible. If $0<\alpha \leq \frac{1}{2}$, then $1=(\alpha-1)\left(\alpha^{4}+\alpha^{3}+1\right)<0$ which again is impossible. We conclude that $\frac{1}{2}<\alpha<2$.

Similarly $\alpha^{4}-\alpha^{2}+2 \alpha \geq 3$ is equivalent to $\alpha^{5}-\alpha^{3}+2 \alpha^{2}-3 \alpha \geq 0$ which is equivalent to $2 \alpha^{2}-4 \alpha+2 \geq 0$. But this is $2(\alpha-1)^{2} \geq 0$ which is valid. Hence $3 \leq \alpha^{6}<4$ and we get $\left[\alpha^{6}\right]=3$.