At the recent IZhO in mathematics there were unusual problems, many of which trolled the majority of participants. P2 was solved by a trick, and many failed to find it. I also did not solve it: during the contest I acted too technically, which did not help. My approach was an attempt to squeeze the desired number between given ones, which is a stronger statement (but, as it turned out, equivalent) that I managed to prove after the contest.

Problem.

Let $$$n$$$ be a positive integer for which there exist positive integers $$$a$$$ and $$$b$$$ such that

Prove that there exists a positive integer $$$c$$$ such that

Official solution

Denote

Adding, we obtain

from which

My solution

Idea.

1. We look for values $$$x_c$$$ that are squeezed between $$$a\sqrt{10}$$$ and $$$b\sqrt{11}$$$;
2. we prove that such a $$$c$$$ exists and is unique;
3. we show that for the remaining $$$c$$$ equality of integer parts is impossible.

Denote

Step 1. Squeezing condition

If

then it immediately follows that

This is equivalent to the inequality

Compute the differences:

Since the denominator is positive, we obtain the equivalent condition

Hence,

where

Step 2. Uniqueness of the integer solution

The length of the segment equals

From the conditions we have

therefore

Thus, at most one integer can lie in $$$[L, R]$$$.

At the same time,

from which

Therefore,

is the unique integer solution; the necessary and sufficient conditions are verified.

Step 3. Absence of other solutions

From the inequality $$$b\sqrt{11} - a\sqrt{10} \lt 1$$$ it follows that

Then

Consequently,

that is,

Similarly,

Hence, no other values of $$$c$$$ exist.