8.3. With three integers, the following was done: one number was decreased by one, one was increased by one, and the third number was squared. As a result, the same set of numbers was obtained. Find these numbers, given that the sum of the numbers is 2013.
Answer: $1007,1006,0$.

Let the numbers in the original set be $a, b, c$. The resulting set will be $a-1, b+1, c^{2}$. From the fact that the sets coincide, we get the equality $a+b+c=(a-1)+(b+1)+c^{2}$. This simplifies to $c=c^{2}$. Therefore, $c=0$ or $c=1$. Since $c^{2}=c$, we get $a-1=b, b+1=a$. This means there are two cases: $b+1, b, 0$ and $b+1, b, 1$. In the first case, we get the sum $2 b+1=2013$, which leads to the answer. In the second case, we get $2 b+2=2013$. This is a contradiction - the left side is even, while the right side is odd.

Comment. An answer without justification - 1 point.