, 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 $c+d \neq 0$ hold. By the second equation we have $b=5$ and therefore $c=-5 (\neq 7)$. 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) $a+b \neq 0$ with $d=9$, $a=-9$.
We therefore have $b+c = d+a = 0$, $a+b \neq 0$, $c+d \neq 0$ 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 $c+d = a+b = 0$, $b+c \neq 0$, $d+a \neq 0$, and this yields $(a, b, c, d) = (3, -3, 7, -7)$.