6. (20 points) Do there exist five pairwise distinct integers such that the sum of any four of them is a square of a natural number?
Solution. We will show a way to construct the desired sequence. Consider the first five squares: $1, 4, 9, 16, 25$ (any five different ones can be taken).

Then the first number is $1 + 4 + 9 + 16 + 25 - 4 \cdot 25 = -45$, the second is $1 + 4 + 9 + 16 + 25 - 4 \cdot 16 = -9$, the third is $1 + 4 + 9 + 16 + 25 - 4 \cdot 9 = 19$, the fourth is $1 + 4 + 9 + 16 + 25 - 4 \cdot 4 = 39$, and the fifth is $1 + 4 + 9 + 16 + 25 - 4 \cdot 1 = 51$. Indeed, if we add any four of them, the sum will be one of the five initial squares, multiplied by 4, which satisfies the condition of the problem. Answer: yes, for example, $-45, -9, 19, 39, 51$.

| Criterion Content | Points |
| :--- | :---: |
| The correct answer is obtained with justification. | 20 |
| There is a logical error or the solution is insufficiently justified in an otherwise correct and justified approach. | 14 |
| The beginning of a possible solution is provided, some intermediate results are obtained, but the further solution is incorrect or absent. | 6 |
| The solution does not meet any of the criteria listed above. | 0 |
| Maximum score | 20 |

## Solution to Variant № 5