Romania — NMO Selection Tests for the Balkan and International Mathematical Olympiads

The required sets are $\{t\}$, $\{-t, t\}$, $\{0, t\}$ and $\{-t, 0, t\}$. It is readily checked that the elements of the Minkowski product of each of these sets and the set $\{-1, 0, 1, 2\}$ form a finite arithmetic progression.

Now, let $A$ and $B$ be sets of real numbers satisfying the conditions in the statement, and let $|A| \ge 2$ (the case $|A| = 1$ is trivial). Clearly, $A$ and $B$ are both finite.

Let $d > 0$ be the difference of the arithmetic progression $AB$, consider two distinct elements of $A$, say $x$ and $x'$, and two distinct elements of $B$, say $y$ and $y'$, and notice that the elements of $A$, respectively $B$, are integral multiples of $d/(y-y')$, respectively $d/(x-x')$. Scaling $A$ and $B$ accordingly, we may (and will) assume that $A$ and $B$ are both sets of integers. Dividing, if necessary, the elements of $A$, respectively $B$, by their greatest common divisor, we may (and will) further assume that the elements of $A$, respectively $B$, are jointly coprime: $\text{gcd } A = 1$ and $\text{gcd } B = 1$. Further, recall that $A$ and $B$ are both finite and let $a^*$, respectively $b^*$, be an element of $A$, respectively $B$, of maximal absolute value. If necessary, multiply by $-1$ to assume $a^* > 0$ and $b^* > 0$. Under these simplifying assumptions, we will show that $A$ is one of the sets $\{-1, 1\}$, $\{0, 1\}$, $\{-1, 0, 1\}$, whence the conclusion.

Since $\text{gcd } B = 1$ and $d$ divides $(x - x')y$ for all $x$ and $x'$ in $A$ and all $y$ in $B$, it follows that $d$ divides the difference of any two members of $A$. Similarly, $d$ divides the difference of any two members of $B$, and since $|B| \ge 4$, it follows that $b^* > d$.

Consider now elements $a$ in $A$ and $b$ in $B$ such that $ab = a^*b^* - d$, and notice that $ab = a^*b^* - d \ge b^* - d > 0$. Moreover, $|a| = a^*$, for otherwise $a^*b^* - d = ab = |a||b| \le (a^* - 1)b^* = a^*b^* - b^* < a^*b^* - d$ which is a contradiction.

This means that $d = a^*b^* - |a||b| = a^*(b^* - |b|) \ge a^*$. Now, since $a^* \le d$, and the elements of $A$ are congruent modulo $d$, the only possible options for $A$ are either subsets of $\{-d, 0, d\}$, or $\{-d/2, d/2\}$ if $d$ is even, or finally sets of the form $\{a^*, a^* - d\}$, where $d > a^* > |a^* - d|$. The first two cases are covered by the answer.

To rule out the last option, notice that $a = a^*$ (since $|a| = a^* > |a^* - d|$), and therefore $d = a^*(b^* - |b|)$. This means that $a^*$ divides $d$, so $a^* \le d/2$ and $|a^* - d| \ge a^*$, in contradiction with $a^* > |a^* - d|$.