Podlipsky o.k.

Oleg drew an empty $50 \times 50$ table and wrote a 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 ("multiplication table"). What is the maximum number of products in this table that could be rational numbers?
Evaluation. Suppose that among the rational numbers there is 0 and it is written at the top side of the table. Let along the left side of the table be written $x$ irrational and $50-x$ rational numbers. Then along the top side are written $50-x$ irrational and $x$ rational numbers. Note that the product of a non-zero rational and an irrational number is irrational. Therefore, in the table, there are at least $x(x-1)+(50-x)^{2} = 2x^{2}-101x+50^{2}$ irrational numbers. The vertex of the parabola $y=2x^{2}-101x+50^{2}$ is at the point 25.25, so the minimum value of this function at an integer point is achieved at $x=25$ and is equal to $25 \cdot 24 + 25^{2} = 1225$. If 0 is replaced by a non-zero rational number, the number of irrational numbers can only increase. Therefore, in the table, there are no more than $2500-1225=1275$ rational numbers.

Example, when there are 1275 rational numbers. Along the left side, place the numbers 1, 2, ..., 24, 25,

$\sqrt{2}, 2\sqrt{2}, \ldots, 25\sqrt{2}$, and along the top - the numbers $0, 26, 27, \ldots, 49, 26\sqrt{2}, 27\sqrt{2}, \ldots, 50\sqrt{2}$. Then

the irrational numbers will be only $25 \cdot 24 + 25^{2} = 1225$ products of non-zero rational and

irrational numbers.

## Answer

1275 products.

## $\underline{\text{Irrational Equations}}$ Methods for solving problems with parameters Problem $\underline{76453}$ Topics: Difficulty: $4-$ Solve the equation $\sqrt{a-\sqrt{a+x}}=x$.

## Solution

Clearly, $a \geq 0, x \geq 0$. Let $y=\sqrt{a+x} \geq 0$, then $x=\sqrt{a-y}$. Squaring and subtracting, we get $x+y=y^{2}-x^{2}$, that is,

$x+y=0$ or $y-x=1$.

$x+y$ can only be zero if $x=y=0$, in which case $a=0$.

If $y=x+1$, then $(x+1)^{2}=x+a, x^{2}+x+1-a=0$. For $a<1$ this equation has no non-negative roots. For $a \geq 1$ the only non-negative root is $\frac{\sqrt{4a-3}-1}{2}$.

## Answer

For $a=0$, $x=0$; for $a \geq 1$, $x=\frac{\sqrt{4a-3}-1}{2}$; for $a<0$ and $0<a<1$, there are no solutions.