Polish MO 2022/1 wrote: Find all real quadruples $(a,b,c,d)$ satisfying the system of equations $$ \left\{ \begin{array}{ll} ab+cd = 6 \\ ac + bd = 3 \\ ad + bc = 2 \\ a + b + c + d = 6. \end{array} \right. $$ Cute problem. We have $(a + d)(b + c) = (ab + cd) + (ac + bd) = 9$ and $(a + d) + (b + c) = 6$. Therefore, we must have $a + d = b + c = 3$. Similarly, $(a + b)(c + d) = 5$ and $(a + b) + (c + d) = 6$. Therefore, $\{ a + b, c + d \} = \{1, 5 \}$. Also similarly, $(a + c)(b + d) = (ab + cd) + (ad + bc) = 8$ and $(a + c) + (b + d) = 6$. Therefore, we must have $\{ a + c, b + d \} = \{ 2, 4 \}$. Note that $2a = (a + b) + (a + c) - (b + c) \in \{ 0, 2, 4, 6 \}$. Therefore, $a \in \{ 0, 1, 2, 3 \}$. However, each of these gives us a unique solution (because each value of $a$ corresponds to a unique value of $a + b$ and $a + c$), and therefore all the solutions are \[ (a,b,c,d) = (0, 1, 2, 3), (1, 0, 3, 2), (2, 3, 0, 1), (3, 2, 1, 0) \]

Hint for anyone who wants to try this

TheMathBob wrote: Find all real quadruples $(a,b,c,d)$ satisfying the system of equations $$ \left\{ \begin{array}{ll} ab+cd = 6 \\ ac + bd = 3 \\ ad + bc = 2 \\ a + b + c + d = 6. \end{array} \right. $$ Take two equation,add them and then do factorization That's it

This problem was proposed by me

$(a+b)(c+d) = ac + ad + bc + bd = 5, (a+b) + (c+d) = 6$ so with making a quadratic equation we have ${(a+b),(c+d)} = {1,5}$. $(a+c)(b+d) = ab + ad + bc + dc = 8, (a+c) + (b+d) = 6$ so with making a quadratic equation we have ${(a+c),(b+d)} = {2,4}$. $(a+d)(b+c) = ab + ac + bd + cd = 9, (a+d) + (b+c) = 6$ so with making a quadratic equation we have ${(a+d),(b+c)} = {3,3}$. Now $(a+b)+(a+c)-(b+c) = {0,4,2,6} \implies a = {0,1,2,3}$. Applying that with other equations gives us $4 $ answers : $(a,b,c,d) = (0, 1, 2, 3), (1, 0, 3, 2), (2, 3, 0, 1), (3, 2, 1, 0)$ I will do one of the quadratic equation parts for more clarification. $(a+b)(c+d) = 5, (a+b) + (c+d) = 6 \implies (a+b) = \frac{5}{(c+d)} \implies \frac{(c+d)^2 + 5}{(c+d)} = 6 \implies (c+d)^2 - 6(c+d) + 5 = 0 \implies ((c+d)-5)((c+d)-1) = 0 \implies (c+d) = {1,5} \implies {(a+b),(c+d)} = {1,5}$.

Adding the first two equations yields $(a+d)(b+c)=9.$ Since $a+d+b+c=6,$ \begin{align*}(a+d)(6-(a+d))=9&\implies (a+d)^2-6(a+d)+9=0\\&\implies (a+d-3)^2=0\end{align*}and $a+d=b+c=3.$ Similarly, $a+c=\{2,4\}$ and $a+b=\{1,5\}.$ Let $a+c=k_1,b+d=6-k_1,a+b=k_2,c+d=6-k_2$ by Vieta. We see $k_1=a+c=a+3-b$ and $k_2=a+b$ so $a=\frac{k_1+k_2-3}{2}$ and $b=\frac{-k_1+k_2+3}{2}.$ Letting $k_1=2,4$ and $k_2=1,5,$ we have $$(a,b,c,d)=(0,1,2,3),(2,3,0,1),(1,0,3,2),(3,1,2,0).$$We can test the solutions to find that they all work. $\square$