249. Find the conditions that the numbers $\alpha_{1}, \alpha_{2}, \alpha_{3}, \alpha_{4}$ must satisfy for the system of 6 equations with 4 unknowns

$$
\left\{\begin{array}{l}
x_{1}+x_{2}=\alpha_{1} \alpha_{2} \\
x_{1}+x_{3}=\alpha_{1} \alpha_{3} \\
x_{1}+x_{4}=\alpha_{1} \alpha_{4} \\
x_{2}+x_{3}=\alpha_{2} \alpha_{3} \\
x_{2}+x_{4}=\alpha_{2} \alpha_{4} \\
x_{3}+x_{4}=\alpha_{3} \alpha_{4}
\end{array}\right.
$$

to have solutions. Find the values of the unknowns $x_{1}, x_{2}$, $x_{3}, x_{4}$.
249. Subtracting the second equation from the first and the sixth from the fifth, and equating the two obtained expressions for \(x_{2}-x_{3}\), we get:

\[
\alpha_{1}\left(\alpha_{2}-\alpha_{3}\right)=\alpha_{4}\left(\alpha_{2}-\alpha_{3}\right), \text{ or }\left(\alpha_{1}-\alpha_{4}\right)\left(\alpha_{2}-\alpha_{3}\right)=0
\]

Similarly, by forming two different expressions for the differences \(x_{1}-x_{2}\) and \(x_{1}-x_{3}\), we obtain two more relations:

\[
\left(\alpha_{1}-\alpha_{2}\right)\left(\alpha_{3}-\alpha_{4}\right)=0, \quad\left(\alpha_{1}-\alpha_{3}\right)\left(\alpha_{2}-\alpha_{4}\right)=0
\]

From the first of the three obtained relations, it follows that \(\alpha_{1}=\alpha_{4}\) or \(\alpha_{2}=\alpha_{3}\). Suppose, for definiteness, that \(\alpha_{2}=\alpha_{3}=\alpha\). From the second relation, it follows that \(\alpha_{1}=\alpha\) or \(\alpha_{4}=\alpha\). Each of these equalities turns the third relation into an identity. Thus, for the system to be consistent, it is necessary that three out of the four quantities \(\alpha_{1}, \alpha_{2}, \alpha_{3}, \alpha_{4}\) be equal to each other.

Now suppose that \(\alpha_{1}=\alpha_{2}=\alpha_{3}=\alpha, \alpha_{4}=\beta\). Recalling the expressions for the differences \(x_{1}-x_{2}, x_{1}-x_{3}, x_{2}-x_{3}\), which we used to obtain the relations between \(\alpha_{1}, \alpha_{2}, \alpha_{3}\) and \(\alpha_{4}\), we immediately get:

\[
x_{1}=x_{2}=x_{3}
\]

Let us now denote \(x_{1}=x_{2}=x_{3}\) simply by \(x\), and \(x_{4}\) by \(y\). In this case, our six equations with four unknowns will reduce to two equations with two unknowns:

\[
2 x=\alpha^{2}, \quad x+y=\alpha \beta
\]

from which

\[
x=\frac{\alpha^{2}}{2} ; \quad y=\alpha\left(\beta-\frac{\alpha}{2}\right)
\]

Note. Similar reasoning can show that a more general system

![](https://cdn.mathpix.com/cropped/2024_05_21_b937612abb2773f7abffg-277.jpg?height=156&width=964&top_left_y=610&top_left_x=179)
consisting of \(C_{n}^{m}\) equations with \(n\) unknowns (\(n>m+1\)), will be solvable only in the following two cases:

\[
\begin{aligned}
& \quad 1^{\circ} \alpha_{1}=\alpha_{2}=\ldots=\alpha_{n-1}=\alpha_{;} \alpha_{n}=\beta \\
& \left(\text { in this case } x_{1}=x_{2}=\ldots=x_{n-1}=\frac{\alpha^{n}}{n}, x_{n}=\alpha^{n-1}\left(\beta-\frac{n-1}{n} \alpha\right)\right) ; \\
& 2^{\circ} n-m+1 \cdot \text { or more of the quantities } \alpha_{1}, \alpha_{2}, \ldots, \alpha_{n} \text { are zero } \\
& \text { (in this case } \left.x_{1}=x_{2}=\ldots=x_{n}=0\right)
\end{aligned}
\]