10.5. Oleg drew an empty $50 \times 50$ table and wrote a non-zero number above each column and to the left of each row. It turned out that all 100 written numbers are distinct, with 50 of them being rational and the other 50 being irrational. Then, in each cell of the table, he wrote the product of the numbers written next to its row and its column (a "multiplication table"). What is the maximum number of products in this table that could be rational numbers? (O. Podlipsky)
Answer: 1250 works.

Solution. First, we show that there are at least 1250 irrational numbers in the table. Suppose along the left side of the table, there are $x$ irrational and $50-x$ rational numbers. Then along the top side, there are $50-x$ irrational and $x$ rational numbers. Since the product of a non-zero rational and an irrational number is always irrational, there are at least $x^{2}+(50-x)^{2}$ irrational numbers in the table. Here, $x^{2}+(50-x)^{2}=2 x^{2}-100 x+50^{2}=2(x-25)^{2}+2 \cdot 25^{2} \geqslant 2 \cdot 25^{2}=1250$, which is what we needed to show. Therefore, there are at most $2500-1250=1250$ rational numbers in the table.

Exactly 1250 rational numbers in the table can occur, for example, in the following case. Along the left side, the numbers are $1,2, \ldots, 24,25, \sqrt{2}, 2 \sqrt{2}, \ldots, 25 \sqrt{2}$, and along the top side, the numbers are $26,27, \ldots, 49,50,26 \sqrt{2}, 27 \sqrt{2}, \ldots, 50 \sqrt{2}$. Then, the only irrational numbers will be the $2 \cdot 25^{2}=1250$ products of a rational and an irrational number.

Comment. Only the correct answer -0 points.

Proof that there are no more than 1250 rational numbers -4 points.

If the proof is only conducted in a special case where an equal number of rational and irrational numbers are along the left (or top) side, 2 points are given instead of 4 for this part of the solution.

The inequality $x^{2}+(50-x)^{2} \geqslant 1250$ (or an equivalent one) is used without proof - 1 point is deducted.

A correct example showing that there could be exactly 1250 rational numbers -3 points.

The statement that the product of a non-zero rational and an irrational number is irrational can be used without proof.