Show that some of the root of the polynomial is not real. \begin{equation*}
p(x)=a_nx^n+a_{n-1}x^{n-1}+\dots+a_3x^3+x^2+x+1.
\end{equation*}

All the coefficients are real. Show that some of  the roots are not real. 

I don't have any idea how to do this, I only have an intuition that something is related to the coefficients of $x^2$ and $x$ and the constant 1.

Kindly help me!!
Adapting the argument from this answer by Noam Elkies.

If $p(x)$ has only real zeros, then so does its reciprocal
$$
x^np(\frac1x)=x^n+x^{n-1}+x^{n-2}+\cdots+a_{n-1}x+a_n.
$$
But by Vieta relations the sum of the roots of this polynomial is $-1$ and the sum of their pairwise products is $1$. Therefore the sum of the squares of the roots is
$$(-1)^2-2\cdot1=-1&lt;0.
$$
Consequently some of the roots must be non-real.
Counterexample: If $n=0$, then $p$ has no roots.