Can all the roots of the polynomial equation (with real coefficients) $a_nx^n+...+a_3x^3+x^2+x+1=0$ be real ? I tried using Vieta's formulae $\prod \alpha=\dfrac {(1)^n}{a_n}$ , $(\prod \alpha )(\sum \dfrac 1 \alpha)=(1)^{n1}\dfrac 1 a_n$ , so $\sum \dfrac1\alpha = 1$ but this is not going anywhere , please help .

I think that [Noam Elkies' answer to a similar question](http://math.stackexchange.com/a/578472/11619) goes through verbatim. Well. Ewan seems to have come up with a very similar idea. +1 – Jyrki Lahtonen Nov 20 '14 at 20:21

@user123733: You have to award the bounty _as well as_ accepting the question (if that's what you want to do). – TonyK Nov 26 '14 at 19:57
2 Answers
The answer is NO. Suppose $P=1+x+x^2+\sum_{k=3}^{n} a_kx^k$ has $n$ real roots. Obviously zero is not a root of $P$, so $Q(x)=x^nP(\frac{1}{x})$ also has $n$ real roots. Let us denote them by $t_1,t_2,\ldots,t_n$ (the $t_i$ are not necessarily distinct).
Since the expansion of $Q(x)$ starts with $x^n+x^{n1}+x^{n2}+\ldots$, we see that
$$ \sum_{i=1}^{n} t_i=(1), \ \sum_{1\leq i < j \leq n}t_it_j=1 \tag{1} $$
But
$$ \begin{array}{lcl} \sum_{1\leq i < j \leq n}(t_it_j)^2 &=& (n1)\sum_{i=1}^n t_i^22\sum_{1\leq i < j \leq n}t_it_j \\ &=& (n1)\left(\sum_{i=1}^n t_i\right)^2 2n\sum_{1\leq i < j \leq n}t_it_j \\ &=& (n1)(2n) \\ &=& (n+1) \end{array} \tag{2} $$
which is impossible. This concludes the proof.
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I think those $(n+1)$'s in equation $(2)$ should be $(n1)$. And then $(2n+4)$ becomes $2n$ etc. But it still works. – TonyK Nov 26 '14 at 19:55

Here is another approach. Rolle's Theorem (a special case of the Mean Value Theorem) states that for a function $f$ continuous on $[a,b]$ and differentiable on $(a,b)$, $f(a) = f(b)$ implies the existence of a $c \in (a,b)$ such that $f(c) = 0$.
We can then say that if a polynomial has $k$ real roots, its derivative must have $k1$ real roots where each root of the derivative lies between the roots of the original polynomial. Now every root $r$ of a polynomial $P(x)$ of degree $n$ corresponds to a root $r' = \frac{1}r$ of $x^nP\left( \frac{1}x \right)$. Thus, $$P(x) = \sum_{k = 3}^n a_kx^k+x^2+x+1$$ has all real roots iff $$G(x) = x^n+x^{n1}+x^{n2} + \sum_{k = 3}^na_kx^{nk}$$ has all real roots. Taking the $n2$th derivative of $G$ gives us that $$\frac{n!}{2}x^2+(n1)!x+(n2)!1$$ has $2$ real roots or $n(n1)x^2+(n1)x+1$ has real roots. However, $$G_{\delta} = (n1)^24n(n1) < 0$$ for all $n > 1$. Thus the result follows.
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