As is known, the variance of a set is the difference between the mean of the squares and the square of the mean. Therefore, the problem can be reformulated: we need to replace one number with two others so that the arithmetic mean and the mean of the squares of the numbers in the set do not change. The arithmetic mean of the given set is 0, so the arithmetic mean of the new set, and therefore the sum of the numbers in it, should also be 0.

In this set, there are 11 numbers, and the sum of the squares is $2 \cdot\left(1^{2}+2^{2}+3^{2}+4^{2}+5^{2}\right)=110$, so the mean of the squares is 10. In the new set, there are 12 numbers, so the sum of the squares of the numbers in the new set should be 120, that is, it increases by 10. Let's replace the number $a$ with the numbers $b$ and $c$. Then $a=b+c$ and

$a^{2}+10=b^{2}+c^{2}$. Therefore, $b^{2}+c^{2}-10=(b+c)^{2}=b^{2}+c^{2}+2 b c$, from which $b c=-5$.

Thus, one of the numbers is 5 or -5, and the other, respectively, -1 or 1. In the first case, $a=4$, in the second case, $a=-4$.

## Answer

We need to replace -4 with 1 and -5 or replace 4 with -1 and 5.

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In the set $-5,-4,-3,-2,-1,0,1,2,3,4,5$ replace one number with two other integers so that the variance of the set and its mean do not change.

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