9.5 What is the largest number of non-overlapping groups into which all integers from 1 to 25 can be divided so that the sum of the numbers in each group is a perfect square?
Solution 1: A group consisting of a single number can only be formed by 5 squares. The remaining 20 numbers must be divided into groups of at least two. Therefore, there will be no more than 15 groups in total. Let's check that exactly 15 groups are not possible. Indeed, in such a case, the numbers 1, 4, 9, 16, 25 form separate groups, and the remaining numbers are divided into pairs. The sum of the remaining numbers is 270, and they can give pairwise sums from 5 to 47. Thus, the number 270 must be the sum of 10 numbers, each of which is 9, 16, 25, or 36. Modulo 9, these numbers give remainders of 0 and 7, so the remainder 7 is used either 0 or 9 times. In the first case, the remainder of the sum modulo 27 is 9, not 0, and in the second case, the sum of all numbers is no more than \(25 \cdot 9 + 36 = 261\).

An example with 14 groups: we will separate the numbers 9, 16, and 25 into individual groups, and the rest will be divided into 11 pairs with a sum of 25.

Solution 2: We can explicitly provide 11 numbers that are not perfect squares and such that their pairwise sums are not squares. These numbers are \(2, 8, 10, 11, 13, 18, 19, 20, 21, 22\), and 24 (this requires extensive manual verification). From this, it follows that the situation with 5 single groups and 10 pairs is impossible.

Solution 3: Suppose the squares form separate groups, and the other numbers are divided into pairs. Then, with the number 21 in a group, there can only be 15, but then in the group with 10, there is only the number 6, and then with the number 19, there is only 17, and we get that a pair for the number 8 cannot be found.

## Criteria:

- Only proven that there are no more than 14 groups, no example provided - 4 points;
- Example provided, but not proven that 15 groups are not possible - 2 points.