We have given four real numbers $a,b,c,d$ s.t. $(1) \;\; a^2 + b^2 + c^2 + d^2 = a + b + c + d - ab = 3$. According to the set of equations (1) we have $3 - 2 \cdot 3 = a^2 + b^2 + c^2 + d^2 - 2(a + b + c + d - ab)$, which is equivalent to $(a + b - 1)^2 + (c - 1)^2 + (d - 1)^2 = 0$, yielding $a + b = c = d = 1$, which inserted in the set of equations (1) give us $3 = a^2 + (1 - a)^2 + 2 = 3 - a(1 - a)$, yielding $a=0$ or $a=1$. Conclusion: The set of equations (1) has exactly two solutions, namely $(a,b,c,d) = (0,1,1,1), (1,0,1,1)$.