# Polynomial function with unimodular coefficients having all roots real

$$\displaystyle f_n(x)=\sum_{i=0}^{n}a_ix^i$$ is a polynomial having $$n$$ real roots where real numbers $$a_0,a_1, a_2,a_3...a_n$$ are unimodular, i.e $$|a_i|=1$$, and $$n>0$$. Then the number of possible values of n is: $$\begin{matrix} a)\ 1 \qquad & a)\ 2 \qquad & a)\ 3 \qquad & a)\ 4 \end{matrix}$$

Since it's an MCQ, I just put $$n=1,2,3$$ and check if they do have all roots real or not.

For $$n=1$$,
$$f(x)=x-1$$ is a possibility.

For $$n=2$$,
$$f(x)=x^2+x-1$$ is a possibility as
$$\Delta = b^2-4ac = 5 > 0$$

For $$n=3$$,
$$f(x)=x^3+x^2-x-1=(x-1)(x+1)^2$$ is a possibility.

and I have no idea how to prove/disprove all the roots real in polynomials with $$n>3$$.

We show $$n \le 3$$ (assuming $$n\ge 2$$ as the problem is trivial for $$n=1$$)
We note that if $$x_k$$ are the roots (at least two), $$x_k \ne 0$$ since the product is $$\pm 1$$ and then $$\Sigma{x_k^2} = 3$$ since it's positive of the form $$1-2(\pm 1)$$. (note that all Newton expressions in the roots, ie sum, symmetric sum of order 2,3.., product are $$\pm 1$$ also)
Applying the same reasoning with the equation with roots $$\frac{1}{x_k}$$ which is obtained by switching the coefficients which sum to power $$n$$ so has same property of all coefficients $$\pm 1$$, we get $$\Sigma{\frac{1}{x_k^2}}=3$$ too, but then we get:
$$9=(\Sigma{x_k^2})(\Sigma{\frac{1}{x_k^2}}) \ge n^2$$ so we are done
• Sorry, I forgot about the $a_0$. Just edited it in. – user42819 May 11 '19 at 14:45