Let $x_1,x_2,x_3,x_3>0$ such that they are all distinct.
Prove that $$(2+x_1+x_2)(2+x_3+x_4)=36$$ $$(2+x_1+x_3)(2+x_2+x_4)=36$$ $$(2+x_1+x_4)(2+x_2+x_3)=36$$
cannot have a solution.
Again note that $x_1,x_2,x_3,x_4$ are distinct positive numbers. Without this assumption, there are solutions to this system.
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