Problem 1515

1. **Proving $S_n$ is false for even $n$:**

Take $a_1 < 0$ and the remaining $a_i = 0$ for $i = 2, 3, \ldots, n$. Then,
$$E_n = (a_1 - 0)(a_1 - 0) \ldots (a_1 - 0) + (0 - a_1)(0 - 0) \ldots (0 - 0) + \ldots + (0 - a_1)(0 - 0) \ldots (0 - 0).$$
Simplifying, we get:
$$E_n = a_1^{n-1} + (n-1)(-a_1) \cdot 0^{n-2} = a_1^{n-1}.$$
Since $a_1 < 0$ and $n-1$ is even, $a_1^{n-1} < 0$. Therefore, $E_n < 0$ for even $n$, and the proposition $S_n$ is false for even $n$.

2. **Proving $S_n$ is false for odd $n \ge 7$:**

Take any $c > a > b$, and let $a_1 = a$, $a_2 = a_3 = a_4 = b$, and $a_5 = a_6 = \ldots = a_n = c$. Then,
$$E_n = (a - b)(a - b)(a - b)(a - c)^{n-4} + (b - a)(b - b)(b - b)(b - c)^{n-4} + \ldots + (c - a)(c - b)(c - b)(c - b).$$
Simplifying, we get:
$$E_n = (a - b)^3 (a - c)^{n-4} + 0 + \ldots + 0.$$
Since $a > b$ and $a < c$, $(a - b)^3 > 0$ and $(a - c)^{n-4} < 0$ for $n \ge 7$. Therefore, $E_n < 0$ for odd $n \ge 7$, and the proposition $S_n$ is false for odd $n \ge 7$.

3. **Proving $S_n$ is true for $n = 3$:**

Assume $a_1 \ge a_2 \ge a_3$. Then,
$$E_3 = (a_1 - a_2)(a_1 - a_3) + (a_2 - a_1)(a_2 - a_3) + (a_3 - a_1)(a_3 - a_2).$$
The sum of the first two terms is non-negative because $a_1 - a_3 \ge a_2 - a_3$. The last term is also non-negative. Hence,
$$E_3 \ge 0.$$
Therefore, the proposition $S_n$ is true for $n = 3$.

4. **Proving $S_n$ is true for $n = 5$:**

Assume $a_1 \ge a_2 \ge a_3 \ge a_4 \ge a_5$. Then,
$$E_5 = (a_1 - a_2)(a_1 - a_3)(a_1 - a_4)(a_1 - a_5) + (a_2 - a_1)(a_2 - a_3)(a_2 - a_4)(a_2 - a_5) + \ldots + (a_5 - a_1)(a_5 - a_2)(a_5 - a_3)(a_5 - a_4).$$
The sum of the first two terms is:
$$(a_1 - a_2)\left[(a_1 - a_3)(a_1 - a_4)(a_1 - a_5) - (a_2 - a_3)(a_2 - a_4)(a_2 - a_5)\right] \ge 0.$$
The third term is non-negative because the first two factors are non-positive and the last two are non-negative. The sum of the last two terms is:
$$(a_4 - a_5)\left[(a_1 - a_5)(a_2 - a_5)(a_3 - a_5) - (a_1 - a_4)(a_2 - a_4)(a_3 - a_4)\right] \ge 0.$$
Hence,
$$E_5 \ge 0.$$
Therefore, the proposition $S_n$ is true for $n = 5$.