$\displaystyle f_n(x)=\sum_{i=0}^{n}a_ix^i$ 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 $n>3$.


1 Answer 1


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

  • $\begingroup$ Sorry, I forgot about the $a_0$. Just edited it in. $\endgroup$
    – user42819
    Commented May 11, 2019 at 14:45
  • $\begingroup$ assumed it as otherwise, the problem is different and makes less sense $\endgroup$
    – Conrad
    Commented May 11, 2019 at 14:46

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .