## Task 3 - 080933

Give all triples of numbers $(a, b, c)$ that satisfy the equations

$$
\begin{array}{ll}
a+b+c=s_{1} & a-b+c=s_{3} \\
a+b-c=s_{2} & a-b-c=s_{4}
\end{array}
$$

under the additional condition that the set of the four numbers $s_{1}, s_{2}, s_{3}, s_{4}$ (regardless of their order) coincides with the set of the four numbers 1, 2, 3, 4!
Addition of the first and fourth equation yields $2 a=s_{1}+s_{4}$.

Addition of the second and third equation yields $2 a=s_{2}+s_{3}$.

This is only possible if $s_{1}=1$ and $s_{4}=4$ or vice versa, and $s_{2}=2, s_{3}=3$ or vice versa. From this it follows that $2 a=5$, so $a=\frac{5}{2}$.

For $s_{1}=1, s_{4}=4$, it then follows that $b+c=-\frac{3}{2}$. Solving for $b$ and substituting into the second equation yields $\frac{5}{2}+c+\frac{3}{2}+c=s_{3}$. If $s_{3}=2$, then $c=-1$, and if $s_{3}=3$, then $c=-\frac{1}{2}$. Substituting into the first equation yields $b=-\frac{1}{2}$ or $b=-1$.

For $s_{1}=4, s_{4}=1$, it follows that $b+c=\frac{3}{2}$. Solving for $b$ and substituting into the second equation yields $\frac{5}{2}+c-\frac{3}{2}+c=s_{3}$. If $s_{3}=2$, then $c=\frac{1}{2}$, and if $s_{3}=3$, then $c=1$. Substituting into the first equation yields $b=1$ or $b=\frac{1}{2}$.

Thus, there are the following possibilities:

$(a, b, c)=\left(\frac{5}{2},-\frac{1}{2},-1\right)\left(\right.$ if $\left.\left(s_{1}, s_{2}, s_{3}, s_{4}=1,3,2,4\right)\right)$
$(a, b, c)=\left(\frac{5}{2},-1,-\frac{1}{2}\right)\left(\right.$ if $\left.\left(s_{1}, s_{2}, s_{3}, s_{4}=1,2,3,4\right)\right)$
$(a, b, c)=\left(\frac{5}{2}, 1, \frac{1}{2}\right)\left(\right.$ if $\left.\left(s_{1}, s_{2}, s_{3}, s_{4}=4,3,2,1\right)\right)$
$(a, b, c)=\left(\frac{5}{2}, \frac{1}{2}, 1\right)\left(\right.$ if $\left.\left(s_{1}, s_{2}, s_{3}, s_{4}=4,2,3,1\right)\right)$
$(a, b, c)=\left(\frac{5}{2}, \frac{1}{2},-1\right)\left(\operatorname{if}\left(s_{1}, s_{2}, s_{3}, s_{4}=2,4,1,3\right)\right)$
$(a, b, c)=\left(\frac{5}{2},-1, \frac{1}{2}\right)\left(\right.$ if $\left.\left(s_{1}, s_{2}, s_{3}, s_{4}=2,1,4,3\right)\right)$
$(a, b, c)=\left(\frac{5}{2}, 1,-\frac{1}{2}\right)\left(\right.$ if $\left.\left(s_{1}, s_{2}, s_{3}, s_{4}=3,4,1,2\right)\right)$
$(a, b, c)=\left(\frac{5}{2},-\frac{1}{2}, 1\right)\left(\right.$ if $\left.\left(s_{1}, s_{2}, s_{3}, s_{4}=3,1,4,2\right)\right)$

and a check confirms that these are indeed solutions.