Roots of $x^n+x+1$

Question

I am trying to prove that $x^n+x+1$ does not have any real roots if n is even, and that it has just one root if n is odd.

My attempt: When n is odd, I can use Bolzano's theorem to prove that there is at least one real root, but I haven't been able to prove that it is unique. I have tried to arrive to a contradiction supposing that there are two roots, but I haven't been able to conclude anything.

As to the case when n is even, I really have no clue.

Could anyone give me a hint?

Thanks in advance.

Answer 1

The extrema occur where

$$nx^{n-1}+1=0.$$

For even $n$, the value at the extremum is

$$\frac1{n^{n/(n-1)}}-\frac1{n^{1/(n-1)}}+1$$ and this is a positive minimum.

For odd $n$, there is no extremum and the function is monotonic and unbounded.

Answer 2

without calculus (except the intuitive intermediate value property for $n$ odd)

$n$ even and $x$ real root,obviously $x <0$ and then for $-1 \le x <0, 1+x \ge 0, x^n >0$ while for $x \le -1, x^n+x \ge 0$ so in either case $x^n+x+1>0$

For $n$ odd as noted above the function is clearly increasing since both $x^n, x$ are increasing functions and it is negative at $-1$ positive at $0$