Austria 2014

We first note that the first equation can be written in the form $a\cdot (b+c)=3\cdot (b+c)$ (and the others analogously)

Case I: $a+b\ne 0$, $b+c\ne 0$, $c+d\ne 0$, $d+a\ne 0$. In this case we have $(a,b,c,d)=(3,5,7,9)$.

Case II: $a+b=b+c=c+d=d+a=0$. In this case we obtain solutions $(a,b,c,d)=(t,-t,t,-t)$, with any real values of $t$.

* Case III: There exists a sum equal to $0$ and there exists a sum not equal to $0$. Let us assume that $b+c=0$ and $c+d\ne 0$ hold. By the second equation we have $b=5$ and therefore $c=-5(\ne 7)$. By the third equation, we therefore have $d+a=0$. There are now two subcases to consider.

Subcase A) $a+b=0$ with $a=-5$, $d=5$ and $c+d=0$, which yields a contradiction.

We therefore have subcase B) $a+b\ne 0$ with $d=9$, $a=-9$. We therefore have $b+c=d+a=0$, $a+b\ne 0$, $c+d\ne 0$ and $(a,b,c,d)=(-9,5,-5,9)$.

Starting with some other pair, analogous reasoning always yields: one sum equal to $0$ and the next (cyclically) not equal to $0$ implies that the one after this is again equal to $0$, and the last again not equal to $0$.

The only other case left is therefore given by $c+d=a+b=0$, $b+c\ne 0$, $d+a\ne 0$, and this yields $(a,b,c,d)=(3,-3,7,-7)$.