Problem 1975

Answer: $-2, 0, 2$.

Call a number $q$ good if every number in the second line appears in the third line unconditionally. We first show that the numbers $0$ and $\pm 2$ are good. The third line necessarily contains $0$, so $0$ is good. For any two numbers $a, b$ in the first line, write $a = x - y$ and $b = u - v$, where $x, y, u, v$ are (not necessarily distinct) numbers on the napkin. We may now write
$$2 a b = 2(x - y)(u - v) = (x - v)^2 + (y - u)^2 - (x - u)^2 - (y - v)^2$$
which shows that $2$ is good. By negating both sides of the above equation, we also see that $-2$ is good.

We now show that $-2, 0$, and $2$ are the only good numbers. Assume for sake of contradiction that $q$ is a good number, where $q \notin \{-2, 0, 2\}$. We now consider some particular choices of numbers on Gugu's napkin to arrive at a contradiction.

Assume that the napkin contains the integers $1, 2, \ldots, 10$. Then, the first line contains the integers $-9, -8, \ldots, 9$. The second line then contains $q$ and $81q$, so the third line must also contain both of them. But the third line only contains integers, so $q$ must be an integer. Furthermore, the third line contains no number greater than $162 = 9^2 + 9^2 - 0^2 - 0^2$ or less than $-162$, so we must have $-162 \leqslant 81q \leqslant 162$. This shows that the only possibilities for $q$ are $\pm 1$.

Now assume that $q = \pm 1$. Let the napkin contain $0, 1, 4, 8, 12, 16, 20, 24, 28, 32$. The first line contains $\pm 1$ and $\pm 4$, so the second line contains $\pm 4$. However, for every number $a$ in the first line, $a \not\equiv 2 \pmod{4}$, so we may conclude that $a^2 \equiv 0, 1 \pmod{8}$. Consequently, every number in the third line must be congruent to $-2, -1, 0, 1, 2 \pmod{8}$; in particular, $\pm 4$ cannot be in the third line, which is a contradiction.

Let $q$ be a good number, as defined in the first solution, and define the polynomial $P\left(x_1, \ldots, x_{10}\right)$ as
$$\prod_{i<j} (x_i - x_j) \prod_{a_i \in S} \left(q(x_1 - x_2)(x_3 - x_4) - (a_1 - a_2)^2 - (a_3 - a_4)^2 + (a_5 - a_6)^2 + (a_7 - a_8)^2\right),$$
where $S = \{x_1, \ldots, x_{10}\}$.

We claim that $P\left(x_1, \ldots, x_{10}\right) = 0$ for every choice of real numbers $\left(x_1, \ldots, x_{10}\right)$. If any two of the $x_i$ are equal, then $P\left(x_1, \ldots, x_{10}\right) = 0$ trivially. If no two are equal, assume that Gugu has those ten numbers $x_1, \ldots, x_{10}$ on his napkin. Then, the number $q(x_1 - x_2)(x_3 - x_4)$ is in the second line, so we must have some $a_1, \ldots, a_8$ so that
$$q(x_1 - x_2)(x_3 - x_4) - (a_1 - a_2)^2 - (a_3 - a_4)^2 + (a_5 - a_6)^2 + (a_7 - a_8)^2 = 0$$
and hence $P\left(x_1, \ldots, x_{10}\right) = 0$.

Since every polynomial that evaluates to zero everywhere is the zero polynomial, and the product of two nonzero polynomials is necessarily nonzero, we may define $F$ such that
$$\begin{equation*} F\left(x_1, \ldots, x_{10}\right) \equiv q(x_1 - x_2)(x_3 - x_4) - (a_1 - a_2)^2 - (a_3 - a_4)^2 + (a_5 - a_6)^2 + (a_7 - a_8)^2 \equiv 0 \tag{1} \end{equation*}$$
for some particular choice $a_i \in S$.

Each of the sets $\{a_1, a_2\}, \{a_3, a_4\}, \{a_5, a_6\}$, and $\{a_7, a_8\}$ is equal to at most one of the four sets $\{x_1, x_3\}, \{x_2, x_3\}, \{x_1, x_4\}$, and $\{x_2, x_4\}$. Thus, without loss of generality, we may assume that at most one of the sets $\{a_1, a_2\}, \{a_3, a_4\}, \{a_5, a_6\}, \{a_7, a_8\}$ is equal to $\{x_1, x_3\}$. Let $u_1, u_3, u_5, u_7$ be the indicator functions for this equality of sets: that is, $u_i = 1$ if and only if $\{a_i, a_{i+1}\} = \{x_1, x_3\}$. By assumption, at least three of the $u_i$ are equal to $0$.

We now compute the coefficient of $x_1 x_3$ in $F$. It is equal to $q + 2(u_1 + u_3 - u_5 - u_7) = 0$, and since at least three of the $u_i$ are zero, we must have that $q \in \{-2, 0, 2\}$, as desired.