<s>[INST] Prove that $x^4 + x^3 + x^2 + x + 1$ has no roots Prove that $x^4 + x^3 + x^2 + x + 1$ has no roots.
I can see two ways of proving it.
The first one is to notice, that $x^5 - 1 = (x - 1)(x^4 + x^3 + x^2 + x + 1)$. It has the only root 1. And it is not the root of $x^4 + x^3 + x^2 + x + 1$. So, $x^4 + x^3 + x^2 + x + 1$ does not have roots.
Another way is to solve it as a palindromic polynomial. It does not have roots.
But is there any way to directly manipulate the expression to show that it is always greater than zero?
 [/INST] Another proof by contradiction.
Suppose $x^4+x^3+x^2+x+1$ has a real root. Necessarily, $x<0$. Set $x=-t,\enspace t>0$. Rewrite the equation as
$$t^4-t^3+t^2-t=t(t-1)(t^2+1)=-1$$
It shows $t-1<0$, hence $\;0<t<1$.
On another hand, $\;t=\dfrac1{(1-t)(1+t^2)}=\dfrac1{1-t+t^2-t^3}$, and since $0<t<1$, we have $0<1-t+t^2-t^3<1-t+t^2=1-t(1-t)<1$, hence
$$t=\dfrac1{1-t+t^2-t^3}>1,\quad\text{contradiction}.$$
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