# Task 8. (16 points)

How many solutions in integers does the equation $x_{1}^{4}+x_{2}^{4}+\cdots+x_{14}^{4}=2031$ have.
# Solution.

Note that if the number $n$ is even, then $n^{4}$ is divisible by 16.

If $n$ is odd, then the number

$$
n^{4}-1=(n-1)(n+1)\left(n^{2}+1\right)
$$

is divisible by 16.

Therefore, the remainder of the left-hand side of the equation $x_{1}^{4}+x_{2}^{4}+\cdots+x_{14}^{4}$ when divided by 16 is equal to the number of odd numbers in the set $x_{1}, x_{2}, \ldots, x_{14}$, i.e., it does not exceed 14. On the other hand, 2031 has a remainder of 15 when divided by 16, which means the equality of the left and right sides is impossible.

Answer: 0 (no solutions).

All-Russian School Olympiad "Mission Possible. Your Calling - Financier!"

| Content of the criterion | Evaluation | Points |
| :---: | :---: | :---: |
| Complete solution. | + | 16 |
| Main logical steps of the solution are presented. Some justifications are missing in the solution. | $\pm$ | 12 |
| The idea of the solution is found, but it is not completed. A significant part of the task is completed. | $+/ 2$ | 8 |
| The solution is generally incorrect or incomplete, but contains some progress in the right direction. In particular, estimates for $x_{1}, x_{2}, \ldots, x_{14}$ and the correct answer may be provided. | $\mp$ | 4 |
| The solution does not meet any of the criteria described above. | $-/ 0$ | 0 |
| Maximum score | 16 |  |