Problem 4. We call a set $M$ of real numbers special if it has the following properties:

(i) for any $x, y \in M, x \neq y$, the numbers $x+y$ and $x y$ are non-zero, and exactly one of them is rational;

(ii) for any $x \in M$, the number $x^{2}$ is irrational.

Find the maximum number of elements in a special set.
Solution. The maximum number requested is 4, an example of a special set with 4 elements being \( M = \{\sqrt{2}-1, \sqrt{2}+1, 2-\sqrt{2}, -2-\sqrt{2}\} \)

2 points

We will show that there are no special sets with at least 5 elements. Clearly, all elements of a special set \( M \) are irrational (it follows from the second condition). We will use the following observations:

O1. If \( x, y, z \) are three distinct elements of \( M \), then \( x+y, x+z \) and \( y+z \) cannot all be rational.

Assuming the contrary, it follows that \( 2(x+y+z) \in \mathbb{Q} \), hence \( x+y+z \in \mathbb{Q} \) and \( x \in \mathbb{Q} \), which is absurd. 1 point

O2. If \( x, y, z \) are three distinct elements of \( M \), then \( xy, xz \) and \( yz \) cannot all be rational.

Assuming the contrary, it follows that \( x^2 yz = (xy) \cdot (xz) \in \mathbb{Q} \) and, since \( yz \in \mathbb{Q} \), it follows that \( x^2 \in \mathbb{Q} \), which is absurd. 1 point

O3. If \( x, y \in M \) and \( xy \in \mathbb{Q} \), then for any \( z \in M \), we have \( x+z \in \mathbb{Q} \) and \( y+z \in \mathbb{Q} \).

Otherwise, from the hypothesis and previous observations, it follows that \( x+z \in \mathbb{Q} \) and \( yz \in \mathbb{Q} \) or \( y+z \in \mathbb{Q} \) and \( xz \in \mathbb{Q} \). In the first case, from \( xy \in \mathbb{Q} \) and \( yz \in \mathbb{Q} \), it follows that \( xy + yz = y(x+z) \in \mathbb{Q} \) and, since \( x+z \in \mathbb{Q} \) and \( x+z \neq 0 \) (from the hypothesis!), it follows that \( y \in \mathbb{Q} \), a contradiction. The other case is identical. 1 point

Now suppose there exists a special set with at least 5 elements \( a, b, c, d, e \). From O1, at least two elements have an irrational sum; let these be \( a \) and \( b \). Then \( ab \in \mathbb{Q} \), and from O3 it follows that \( a+c, a+d, a+e \) are rational. According to O1, the numbers \( c+d, c+e \) and \( d+e \) cannot be rational, and from the hypothesis it follows that \( cd \) and \( de \) are rational, which contradicts O2. 2 points