International Mathematical Olympiad

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 $$2ab=2(x-y)(u-v)=(x-v{)}^2+(y-u{)}^2-(x-u{)}^2-(y-v{)}^2$$ 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 $1,2,\dots ,10$. Then, the first line contains the integers $-9,-8,\dots ,9$. 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 $-162⩽81q⩽162$. 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, $a\not\equiv 2(\mathrm{mod}4)$, so we may conclude that $a^2\equiv 0,1(\mathrm{mod}8)$. Consequently, every number in the third line must be congruent to $-2,-1,0,1,2(\mathrm{mod}8)$; in particular, $\pm 4$ cannot be in the third line, which is a contradiction.

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Alternative solution.

Let $q$ be a good number, as defined in the first solution, and define the polynomial $P\left(x_1,\dots ,x_{10}\right)$ 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 $S=\{x_1,\dots ,x_{10}\}$.

We claim that $P\left(x_1,\dots ,x_{10}\right)=0$ for every choice of real numbers $\left(x_1,\dots ,x_{10}\right)$. If any two of the $x_i$ are equal, then $P\left(x_1,\dots ,x_{10}\right)=0$ trivially. If no two are equal, assume that Gugu has those ten numbers $x_1,\dots ,x_{10}$ 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 $a_1,\dots ,a_8$ 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 $P\left(x_1,\dots ,x_{10}\right)=0$.

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 $$\begin{array}{c}F\left(x_1,\dots ,x_{10}\right)\equiv q(x_1-x_2)(x_3-x_4)-(a_1-a_2{)}^2-(a_3-a_4{)}^2+(a_5-a_6{)}^2+(a_7-a_8{)}^2\equiv 0\end{array}\text{\hspace{1em}(1)}$$ 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.