silk road special polynomials, one subset with no real root, one with real roots

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silk road special polynomials, one subset with no real root, one with real roots

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Source: SRMC 2007

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#1 h Sep 2, 2018, 9:08 PM • 1 Y

Y by Adventure10

The set of polynomials $f_1, f_2, \ldots, f_n$ with real coefficients is called special , if for any different $i,j,k \in \{ 1,2, \ldots, n\}$ polynomial $\dfrac{2}{3}f_i + f_j + f_k$ has no real roots, but for any different $p,q,r,s \in \{ 1,2, \ldots, n\}$ of a polynomial $f_p + f_q + f_r + f_s$ there is a real root.
a) Give an example of a special set of four polynomials whose sum is not a zero polynomial.
b) Is there a special set of five polynomials?

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#2 h Sep 20, 2019, 12:25 AM • 2 Y

Y by Adventure10, Mango247

Here's a solution for part (b). Assume the contrary, for contradiction.

Let $d_1, d_2, d_3, d_4, d_5$ be the degrees of $f_1, f_2, \cdots, f_5$ respectively, where $d_1 \ge d_2 \ge \cdots \ge d_5$ WLOG. We will conduct casework on the number of the $d_i$ 's which are equal to $d_1$ .

Case 1. $1$

Then observe that $\frac23 f_1 + f_2 + f_3, \frac23 f_1 + f_2 + f_4, \frac23 f_1 + f_3 + f_4$ must all be always positive or always negative, and so their sum cannot have real roots. However, they sum to $2(f_1 + f_2 + f_3 + f_4)$ , so we lose.

Case 2. $2$

Do the same thing, but on $\frac23 f_1 + f_3 + f_4, \frac23 f_1 + f_3 + f_5, \frac23 f_1 + f_4 + f_5.$

Case 3. $3$

Let $f_1, f_2, f_3$ have leading coefficients of $a_1, a_2, a_3$ . WLOG let $a_1 \ge a_2 \ge a_3$ . WLOG let $a_1, a_2 > 0$ (else negate all the polynomials). Notice that $\frac23 f_1 + f_2 + f_3, \frac23 f_1 + f_2 + f_4, \frac23 f_1 + f_3 + f_4$ all must take only positive values on the reals, and hence so must their sum. This, however, means that $f_1 + f_2 + f_3 + f_4$ has no roots.

Case 4. $4$ .

Let $f_1, f_2, f_3, f_4$ have leading coefficients of $a_1, a_2, a_3, a_4$ , where we WLOG assume $a_1 \ge a_2 \ge a_3 \ge a_4.$ If $a_2 + a_3 < 0$ , then observe that $\frac23 f_5 + f_2 + f_3, \frac23 f_5 + f_2 + f_4, \frac23 f_5 + f_3 + f_4$ all take only negative values, and so summing gives a contradiction. If $a_2 + a_3 >0$ , then $\frac23 f_5 + f_1 + f_2, \frac23 f_5 + f_1 + f_3, \frac23 f_5 + f_2 + f_3$ take only positive values. Now, let's examine the case when $a_2 + a_3 = 0.$ Now, observe that $\frac23 f_1 + f_2 + f_3, \frac23 f_1 + f_2 + f_5, \frac23 f_1 + f_3 + f_5$ all take only positive values, and so summing gives another contradiction.

Case 5. $5$ .

Let $f_1, f_2, f_3, f_4, f_5$ have leading coefficients of $a_1, a_2, a_3, a_4, a_5$ respectively, where we WLOG assume that $a_1 \ge a_2 \ge a_3 \ge a_4 \ge a_5.$ If $\frac23 a_4 + a_2 + a_3 > 0$ , then we are done by summing $\frac23 f_4 + f_2 + f_3, \frac23 f_4 + f_1 + f_3, \frac23 f_4 + f_1 + f_2$ . We can finish similarly if $\frac23 a_2 + a_3 + a_4 < 0.$ Since $\frac23 a_2 + a_3 + a_4 \le \frac23 a_4 + a_2 + a_3$ , the only possible case where we are not done is if $\frac23 a_2 + a_3 + a_4 = \frac23 a_4 + a_2 + a_3 = 0$ . However, subtracting implies that $a_2 = a_4 \Rightarrow a_2 = a_3 = a_4 = 0$ . This is a contradiction, since leading coefficients are not $0$ by definition.

As we've exhausted all cases, we've shown that no special set of five polynomials exists.

$\square$

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