<s>[INST] Number of real roots $x^8-x^5+x^2-x+1=0$ Find the number of real roots of $x^8-x^5+x^2-x+1$.
My attempt: $f(x)$ has $4$ sign changes and $f(-x)$ has no sign changes, so the possibility of having real roots is $4+0=4$. Since this is a polynomial of degree $8$ it should have $8-4=4$ imaginary roots.
It is quite impossible to see that this equation does not have any real roots by observing the factor $x^8-x^5+x^2-x+1= (x-1)(x^7+x^6+x^5+x)+1>0$, hence it can't have any real roots. My question is how to prove using this equation doesn't have any real roots using Descartes' Rule? Please give some useful hints.
 [/INST] Another way: note that
$$
x^8-x^5+x^2-x+1=\left(x^4-\frac{1}{2}x\right)^2+\frac{1}{2}x^2+\left(\frac{1}{2}x-1\right)^2>0.
$$
Equality cannot occur since $x^4-\frac{1}{2}x$, $x$ and $\frac{1}{2}x-1$ cannot vanish together.
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