Let $k\geq2$ be even and let $f(x)=x^{k}+x^{k-1}+...+1\in\mathbb{Q}[x]$
I want to prove that there is no linear polynomial that divides $f(x)$
So I figured that if there was $g(x)=x-\alpha$ that divides $f(x)$ I would should've get $f(\alpha)=0$, so I need to show that $f(\alpha)\neq0$...
Any hints? Plus, if someone can pinpoint the difference between $\mathbb{R}[x]$ and $\mathbb{Q}[x]$ in this proof will be great!
On
Note that $f(x)$ is monic with integer coefficients, and hence, any root of $f(x)$ is an algebraic integer. Recall that if an algebraic integer is rational, then it is an integer. Hence, it suffices to show that $f(x)$ has no roots in $\mathbb{Z}$. Let $\alpha \in \mathbb{Z}$. It is clear $\alpha \geq 0$ is not a root, so we consider $\alpha < 0$. To see that $f(-1) \neq 0$, reduce modulo $2$, and note that $$f(-1) = (-1)^{2k} + (-1)^{2k-1} + \cdots + 1 \equiv 1^{2k} + 1^{2k-1} + \cdots + 1 \equiv \underbrace{1+1+\cdots+1}_{2k+1\text{ times}} \equiv 1 \pmod 2$$
Consider $\alpha < -1$. Then $|\alpha| > 1$, so in particular, $|\alpha|^{n} < |\alpha|^{n+1}$ for any natural $n$. We see then that:
$$f(\alpha) = (\alpha^{2k}+\alpha^{2k-1})+(\alpha^{2k-2}+\alpha^{2k-3})+\cdots+(\alpha^{2}+\alpha)+1$$
But each term $(\alpha^{2i}+\alpha^{2i-1}) > 0$, so $f(\alpha) > 0$ for every integer $\alpha < -1$. Hence, $f(x)$ has no roots over the integers and hence no roots over the rationals.
Alternatively, you can again reduce $f(\alpha)$ modulo $2$; in either case, you can see $f(k) \equiv 1 \pmod 2$.
With the caveat of AWertheim ($(x+1)$ divides $f(x)$ for $k$ odd), hint: what is and what roots in $\Bbb C$ has $(x-1)f(x)$?