## Task 2 - 221242

Determine all real numbers $a$ for which there exist nonnegative real numbers $x_{1}, x_{2}, x_{3}, x_{4}$ that satisfy the following equations (1), (2), (3), (4):

$$
\begin{aligned}
x_{1}+x_{2}+x_{3}+x_{4} & =1 \\
x_{1}+2 x_{2}+3 x_{3}+4 x_{4} & =a \\
x_{1}+2^{2} x_{2}+3^{2} x_{3}+4^{2} x_{4} & =a^{2} \\
x_{1}+2^{3} x_{2}+3^{3} x_{3}+4^{3} x_{4} & =a^{3}
\end{aligned}
$$
First, it is about solving the above linear equation system in $x_{1}, x_{2}, x_{3}, x_{4}$. For this, there is a nice trick due to its special form, which I will demonstrate explicitly here using the calculation of $x_{1}$ as an example:

$$
\begin{array}{r}
f_{1}(a):=(a-2)(a-3)(a-4)=a^{3}-9 a^{2}+26 a-24= \\
=\left(x_{1}+2^{3} x_{2}+3^{3} x_{3}+4^{3} x_{4}\right)-9\left(x_{1}+2^{2} x_{2}+3^{2} x_{3}+4^{2} x_{4}\right)+26\left(x_{1}+2 x_{2}+3 x_{3}+4 x_{4}\right)- \\
-24\left(x_{1}+x_{2}+x_{3}+x_{4}\right)=f_{1}(1) x_{1}+f_{1}(2) x_{2}+f_{1}(3) x_{3}+f_{1}(4) x_{4}=f_{1}(1) x_{1}
\end{array}
$$

and thus

$$
x_{1}=\frac{1}{f_{1}(1)}(a-2)(a-3)(a-4)=-\frac{1}{6}(a-2)(a-3)(a-4)
$$

In a similar way, using the polynomial expressions

$$
f_{2}(a):=(a-1)(a-3)(a-4), f_{3}(a):=(a-1)(a-2)(a-4), f_{4}(a):=(a-1)(a-2)(a-3)
$$

in $a$, the values of the other variables are obtained as

$$
\begin{aligned}
& x_{2}=\frac{1}{f_{2}(2)}(a-1)(a-3)(a-4)=\frac{1}{2}(a-1)(a-3)(a-4) \\
& x_{3}=\frac{1}{f_{3}(3)}(a-1)(a-2)(a-4)=-\frac{1}{2}(a-1)(a-2)(a-4) \\
& x_{4}=\frac{1}{f_{4}(4)}(a-1)(a-2)(a-3)=\frac{1}{6}(a-1)(a-2)(a-3)
\end{aligned}
$$

From this, it can be immediately seen by substitution that for $a \notin\{1,2,3,4\}$, at least one of the 4 variables is negative, while for $a \in\{1,2,3,4\}$, each of the 4 values of $a$ results in 3 of the variables $x_{1}, x_{2}, x_{3}, x_{4}$ being 0, and the fourth being positive, i.e., exactly these four values of $a$ represent the solution to the problem here.