Find all quadruples $(a, b, c, d)$ of real numbers for which

$$
\begin{aligned}
& a b+c+d=3, \\
& b c+d+a=5, \\
& c d+a+b=2, \\
& d a+b+c=6 .
\end{aligned}
$$
Subtracting the first two equations from each other gives

$$
2=5-3=(b c+d+a)-(a b+c+d)=b(c-a)+a-c=(b-1)(c-a) .
$$

Subtracting the last two equations from each other gives

$$
4=6-2=(d a+b+c)-(c d+a+b)=d(a-c)+c-a=(1-d)(c-a)
$$

We see that $c-a$ is not equal to 0 (because then the product of $c-a$ with something else would have to be 0), and we can thus conclude that $1-d=2(b-1)$, or $3=2 b+d$. In the same way, we combine the second and third equations; this yields $3=(c-1)(b-d)$. Combining the first and fourth equations gives $3=(1-a)(b-d)$. Therefore, $c-1=1-a$, or $a+c=2$.
Now we add the first two equations together. This gives
$8=3+5=a b+c+d+b c+d+a=b(a+c)+(a+c)+2 d=2 b+2+2 d=2+3+d=5+d$,
so $d=3$. From $2 b+d=3$, it follows that $b=0$. The first equation now becomes $0+c+3=3$, so $c=0$. In the second equation, we finally find $0+3+a=5$, so $a=2$. We have now found that the only possible solution is: $(a, b, c, d)=(2,0,0,3)$. Checking shows that this quadruple indeed satisfies all four equations.