Answer: No.

Solution: Suppose such numbers exist, denote them as $x-2, x-1, x, x+1, x+2$ for some natural number $x \geq 3$. Notice that the ten numbers in parentheses on both sides of the equation in the problem are all possible pairwise sums of the numbers $a, b, c, d, e$, that is, the pairwise sums of the numbers $x-2, x-1, x, x+1, x+2$. From the equality in the problem, it follows that the product of all these ten pairwise sums is a perfect square of a natural number. Express this product in terms of $x: \left(4 x^{2}-9\right)\left(4 x^{2}-4\right)\left(4 x^{2}-1\right)^{2} 4 x^{2}$, it is a square if and only if $\left(4 x^{2}-9\right)\left(4 x^{2}-4\right)=16 x^{4}-52 x^{2}+36$ is a square. However, the last expression cannot be a square, as it is less than $\left(4 x^{2}-6\right)^{2}=16 x^{4}-48 x^{2}+36$, but greater than $\left(4 x^{2}-7\right)^{2}=16 x^{4}-56 x^{2}+49$, due to the fact that $4 x^{2} \geq 36>13$.

Grading Criteria. Noted that the ten numbers in parentheses on both sides of the equation in the problem are all possible pairwise sums of the numbers $a, b, c, d, e: 1$ point. Noted that the product of all these ten pairwise sums is a perfect square of a natural number: 2 points. This product is expressed in terms of $x$, and noted that it is a square if and only if $16 x^{4}-52 x^{2}+36$ is a square: 2 points. Proved that $16 x^{4}-52 x^{2}+36$ is not a square: 2 points.

## Criteria for Determining Winners and Prize Winners of the All-Siberian Open School Olympiad in Mathematics (2015-2016 academic year)

According to the Regulations, the winners and prize winners of the Olympiad were determined based on the results of the Final Stage of the Olympiad. The total number of winners and prize winners was 380 out of 1578 participants, which is $24.08 \%$. The number of winners was 85, which is $5.38 \%$.

Based on the overall ranking of participants and taking into account the noticeable gaps in the scores of the groups of participants at the top of the ranking, the jury of the Olympiad developed the following criteria for determining winners and prize winners: The maximum possible number of points - 35 points.

## 11th Grade:

Winners:

Participants who scored more than $77 \%$ of the maximum number of points, i.e., from 27 to 35 points; Prize winners:

2nd degree - more than $62 \%$ of the maximum number of points, i.e., from 22 to 26 points

3rd degree - more than $51 \%$ of the maximum number of points, i.e., from 18 to 21 points

10th Grade:

Winners:

Participants who scored more than $85 \%$ of the maximum number of points, i.e., from 30 to 35 points;

Prize winners:

2nd degree - more than $62 \%$ of the maximum number of points, i.e., from 22 to 29 points

3rd degree - more than $51 \%$ of the maximum number of points, i.e., from 18 to 21 points

9th Grade:

Winners:

Participants who scored more than $85 \%$ of the maximum number of points, i.e., from 30 to 35 points;

Prize winners:

2nd degree - more than $65 \%$ of the maximum number of points, i.e., from 24 to 29 points

3rd degree - more than $51 \%$ of the maximum number of points, i.e., from 18 to 23 points

8th Grade:

Winners:

Participants who scored more than $82 \%$ of the maximum number of points, i.e., from 29 to 35 points;

Prize winners:

2nd degree - more than $62 \%$ of the maximum number of points, i.e., from 22 to 28 points

3rd degree - more than $48 \%$ of the maximum number of points, i.e., from 17 to 21 points

7th Grade:

Winners:

Participants who scored more than $85 \%$ of the maximum number of points, i.e., from 30 to 35 points;

Prize winners:

2nd degree - more than $71 \%$ of the maximum number of points, i.e., from 24 to 29 points

3rd degree - more than $42 \%$ of the maximum number of points, i.e., from 15 to 21 points

Co-Chair of the Mathematics Jury

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A.Yu. Avdyushenko