2. Find all such triples of numbers $m, n, k$, such that each of the equations $m x^{2}+n=0, n x^{2}+k=0$ and $k x^{2}+m=0$ has at least one solution.
Answer: $m=n=k=0$. Solution. If $m=0$, then from $m x^{2}+n=0$ we get $n=0$, and from $n x^{2}+k=0$ we get $k=0$. Similarly for $n=0$ and $k=0$. Thus, if one of the numbers $m, n, k$ is 0, then the other two are also 0. Suppose none of the numbers $m, n, k$ are 0. Then from the first equation it follows that the numbers $m$ and $n$ have different signs, and from the second that the numbers $n$ and $k$ have different signs. But then the numbers $m$ and $k$ have the same sign, and in the third equation $x^{2}=-m / k<0$, meaning it has no solutions. The obtained contradiction shows that the numbers $m, n$, and $k$ cannot all be non-zero. Note. The fact that each of the equations specified in the condition has a solution does not mean that all three equations have the same solution. Therefore, proving that the system of three given equations has a solution only when $m=n=k=0$ is not a solution to the problem.

Instructions. Only the answer - 0 points. Answer plus proof that if one of the numbers $m, n, k$ is 0, then the others are also 0 - 1 point. It is shown that all three numbers cannot be non-zero, but it is not shown that if one of the numbers $m, n, k$ is 0, then the others are also 0 (usually this solution assumes that $x^{2}$ is always positive) - 4 points. Interpreting the problem as the study of a system of three given equations - a gross error, 0 points.

Attention! Due to technical reasons, those who solved problem 2 in their personal accounts were given a condition in which the symbol $x$ was missing: Find all such triples of numbers $m, n, k$ such that each of the equations $m^{2}+n=0, n^{2}+k=0$, and $k^{2}+m=0$ has at least one solution. In this formulation, the problem retains its meaning but becomes significantly easier, as it now indeed reduces to solving a system of three given equations, from which we get the equality $k=-n^{2}=-m^{4}=-k^{8}$, hence either $k=m=n=0$ or $k=m=n=-1$. For those who solved the problem in this formulation, the actually solved problem is evaluated. Their results are recorded on a separate sheet "Problem without x" in the results table. Results of those who solved problem 2 in the correct formulation or did not write its solution are recorded in the sheet "Problem with x". If both formulations of the problem are solved in the work, record the results in the sheet corresponding to the solution that is evaluated higher.

Instructions for the problem without x. Only the answer - 0 points. No check of the answer if it is necessary during the solution - 1 point is deducted. Due to division by an expression that can be 0, the solution $k=m=n=0$ is lost, the rest is correct - 3 points. Generally, it is not taken into account that $k, m, n$ can be zeros - no more than 3 points. It is shown that $m, n, k$ can only be 0 or -1, but it is not shown that $m=n=k$ - 4 points. The equation $m^{8}+m=0$ is formed, further substantial progress is absent - 1 point. The same plus it is indicated that the roots of this equation are 0 and -1 (but it is not said that $k=m=n$) - 3 points. The equation $m^{8}+m=0$ is formed, its roots are indicated, it is said but not proven that $k=m=n$ - 4 points.