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Find all polynomials with coefficients from set $\{-1,1\}$ and which have all their roots real.

What I have tried:

Assume polynomial is $a_{n}x^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\cdots +a_{0}=0$ and $a_{i}\in \{-1,1\}$ for $i=0,1,2,3,4,\cdots ,n$. Let its roots be $x=r_{i}$ for $i=1,2,3,\cdots n$. Then

$\begin{aligned}\sum r_{i}=-\frac{a_{n-1}}{a_{n}}=\pm 1\\\\ \mathop{\sum}_{i<j}r_{i}r_{j}=+\frac{a_{n-2}}{a_{n}}=\pm 1\\\\ \mathop{\sum}_{i<j<k}r_{i}r_{j}r_{k}=+\frac{a_{n-2}}{a_{n}}=\pm 1\\\\ \prod r_{i}=(-1)^n\frac{a_{0}}{a_{n}}\end{aligned}$

How do I solve it?

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  • $\begingroup$ Are you actually required to use that form? It'd be a lot easier - to the point of triviality - to use the roots form $a\product_k(x-x_k)$ $\endgroup$ – Dan Uznanski Mar 16 at 9:00
  • $\begingroup$ Note if $a_n = -1$, you can multiply the equation by $-1$ and still get the same roots. Thus, WLOG, you can set $a_n = 1$, with this at least simplifying your equations slightly. Also, your triple product formula should have $-\frac{a_{n-3}}{a_n}$ in the middle part. $\endgroup$ – John Omielan Mar 16 at 9:09
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    $\begingroup$ Have you tried solving the problems explicitly for low degrees and looking for patterns that you can try to prove? $\endgroup$ – Henning Makholm Mar 16 at 9:48
  • $\begingroup$ I haven't solved it, but for $n \ge 2$, I have determined so far that, if you let $a_n = 1$, then $a_{n-2} = -1$, plus that $\sum r_i^2 = 3$. Are you able to see why this is so? $\endgroup$ – John Omielan Mar 16 at 10:30
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WLOG, let $a_n = 1$. Also, let $r_1,\ldots,r_n$ be the roots of the polynomial.

As you noted, $\displaystyle\sum_{k = 1}^{n}r_k = -a_{n-1}$, $\displaystyle\sum_{1 \le k < \ell \le n}r_kr_{\ell} = a_{n-2}$, and $\displaystyle\prod_{k = 1}^{n}r_k = (-1)^na_0$. Hence,

$$\displaystyle\sum_{k = 1}^{n}r_k^2 = \left(\sum_{k = 1}^{n}r_k\right)^2-2\left(\sum_{1 \le k < \ell \le n}r_kr_{\ell}\right) = a_{n-1}^2-2a_{n-2} = 1-2a_{n-2} = \begin{cases}3 & \text{if} \ a_{n-2} = -1 \\ -1 & \text{if} \ a_{n-2} = 1 \end{cases}.$$

Since all the roots are real, $\displaystyle\sum_{k = 1}^{n}r_k^2 \ge 0$. Hence, we must have $a_{n-2} = -1$ and thus $\displaystyle\sum_{k = 1}^{n}r_k^2 = 3$.

Trivially, we then have $\displaystyle\sum_{k = 1}^{n}|r_k|^2 = \sum_{k = 1}^{n}r_k^2= 3$, as well as $\displaystyle\prod_{k = 1}^{n}|r_k| = \left|\prod_{k = 1}^{n}r_k\right| = \left|(-1)^na_0\right| = 1$.

So by the RMS-GM inequality, we have $$\displaystyle 1 = \left(\prod_{k = 1}^{n}|r_k|\right)^{1/n} \le \left(\dfrac{1}{n}\sum_{k = 1}^{n}|r_k|^2\right)^{1/2} = \sqrt{\dfrac{3}{n}}.$$

Therefore, we must have $n \le 3$. It is easy to check that the only solutions are:

$x-1$, $x+1$, $x^2-x-1$, $x^2+x-1$, $x^3-x^2-x+1$, $x^3+x^2-x-1$, and their negatives.

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