Problem 1490

Let's break down the solution into detailed steps for clarity.

### Part (a)
We need to find an example of a special set of four polynomials whose sum is not a zero polynomial.

1. Choose Polynomials:
Consider the polynomials:
$$f_1(x) = x^2 + 1, \quad f_2(x) = x^2 + 2, \quad f_3(x) = x^2 + 3, \quad f_4(x) = x^2 + 4$$

2. Check the Condition for Three Polynomials:
For any different $i, j, k \in \{1, 2, 3, 4\}$, we need to check that $\frac{2}{3}f_i + f_j + f_k$ has no real roots.
$$\frac{2}{3}f_i(x) + f_j(x) + f_k(x) = \frac{2}{3}(x^2 + a_i) + (x^2 + a_j) + (x^2 + a_k)$$
Simplifying, we get:
$$\frac{2}{3}x^2 + \frac{2}{3}a_i + x^2 + a_j + x^2 + a_k = \left(\frac{2}{3} + 1 + 1\right)x^2 + \frac{2}{3}a_i + a_j + a_k = \frac{8}{3}x^2 + \frac{2}{3}a_i + a_j + a_k$$
Since $\frac{8}{3}x^2$ is always positive for all $x \neq 0$, and $\frac{2}{3}a_i + a_j + a_k$ is a positive constant, $\frac{2}{3}f_i + f_j + f_k$ has no real roots.

3. Check the Condition for Four Polynomials:
For any different $p, q, r, s \in \{1, 2, 3, 4\}$, we need to check that $f_p + f_q + f_r + f_s$ has a real root.
$$f_p(x) + f_q(x) + f_r(x) + f_s(x) = (x^2 + a_p) + (x^2 + a_q) + (x^2 + a_r) + (x^2 + a_s) = 4x^2 + a_p + a_q + a_r + a_s$$
Since $4x^2$ is always positive for all $x \neq 0$, and $a_p + a_q + a_r + a_s$ is a positive constant, $f_p + f_q + f_r + f_s$ has a real root.

Thus, the set $\{f_1, f_2, f_3, f_4\}$ is a special set of four polynomials.

### Part (b)
We need to prove that there is no special set of five polynomials.

1. Assume the Contrary:
Assume there exists a special set of five polynomials $f_1, f_2, f_3, f_4, f_5$ with degrees $d_1, d_2, d_3, d_4, d_5$ such that $d_1 \ge d_2 \ge d_3 \ge d_4 \ge d_5$.

2. **Case 1: $d_1$ is Unique:**
If $d_1$ is unique, consider $\frac{2}{3}f_1 + f_2 + f_3$, $\frac{2}{3}f_1 + f_2 + f_4$, and $\frac{2}{3}f_1 + f_3 + f_4$. These must all be always positive or always negative, and so their sum cannot have real roots. However, they sum to $2(f_1 + f_2 + f_3 + f_4)$, which contradicts the condition that $f_1 + f_2 + f_3 + f_4$ has a real root.

3. **Case 2: Two Polynomials with Degree $d_1$:**
If two polynomials have degree $d_1$, consider $\frac{2}{3}f_1 + f_3 + f_4$, $\frac{2}{3}f_1 + f_3 + f_5$, and $\frac{2}{3}f_1 + f_4 + f_5$. These must all be always positive or always negative, and so their sum cannot have real roots. However, they sum to $2(f_1 + f_3 + f_4 + f_5)$, which contradicts the condition that $f_1 + f_3 + f_4 + f_5$ has a real root.

4. **Case 3: Three Polynomials with Degree $d_1$:**
Let $f_1, f_2, f_3$ have leading coefficients $a_1, a_2, a_3$. WLOG, let $a_1 \ge a_2 \ge a_3$. WLOG, let $a_1, a_2 > 0$ (else negate all the polynomials). Notice that $\frac{2}{3}f_1 + f_2 + f_3$, $\frac{2}{3}f_1 + f_2 + f_4$, $\frac{2}{3}f_1 + f_3 + f_4$ 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.

5. **Case 4: Four Polynomials with Degree $d_1$:**
Let $f_1, f_2, f_3, f_4$ have leading coefficients $a_1, a_2, a_3, a_4$, where we WLOG assume $a_1 \ge a_2 \ge a_3 \ge a_4$. If $a_2 + a_3 < 0$, then observe that $\frac{2}{3}f_5 + f_2 + f_3$, $\frac{2}{3}f_5 + f_2 + f_4$, $\frac{2}{3}f_5 + f_3 + f_4$ all take only negative values, and so summing gives a contradiction. If $a_2 + a_3 > 0$, then $\frac{2}{3}f_5 + f_1 + f_2$, $\frac{2}{3}f_5 + f_1 + f_3$, $\frac{2}{3}f_5 + f_2 + f_3$ take only positive values. Now, let's examine the case when $a_2 + a_3 = 0$. Now, observe that $\frac{2}{3}f_1 + f_2 + f_3$, $\frac{2}{3}f_1 + f_2 + f_5$, $\frac{2}{3}f_1 + f_3 + f_5$ all take only positive values, and so summing gives another contradiction.

6. **Case 5: All Five Polynomials with Degree $d_1$:**
Let $f_1, f_2, f_3, f_4, f_5$ have leading coefficients $a_1, a_2, a_3, a_4, a_5$ respectively, where we WLOG assume that $a_1 \ge a_2 \ge a_3 \ge a_4 \ge a_5$. If $\frac{2}{3}a_4 + a_2 + a_3 > 0$, then we are done by summing $\frac{2}{3}f_4 + f_2 + f_3$, $\frac{2}{3}f_4 + f_1 + f_3$, $\frac{2}{3}f_4 + f_1 + f_2$. We can finish similarly if $\frac{2}{3}a_2 + a_3 + a_4 < 0$. Since $\frac{2}{3}a_2 + a_3 + a_4 \le \frac{2}{3}a_4 + a_2 + a_3$, the only possible case where we are not done is if $\frac{2}{3}a_2 + a_3 + a_4 = \frac{2}{3}a_4 + a_2 + a_3 = 0$. 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.

$\blacksquare$