Let four functions are f1,f2,f3,f4, fi=x^2＋ai・x＋bi.If y=x^2＋ax＋b and y=x^2＋a'x＋b' intersect at one point, a≠a'.And if intersect no point, a=a' and b≠b'. Without loss of generality, f1 and f2, f1 and f3 intersect one point.So, f4 intersects f1 or f2 and f3. Case1)f4 intersects f1 at one point Without loss of generality,we suppose that f2 and f3 intersect at one point.Since f2 and f4, f3 and f4 intersect no point, a2=a3=a4 which is absurd. Case2)f4 intersects f2 and f3 at one point Since f1 and f4 intersect no point, a1=a4=s and b1≠b4.Similarly,a2=a3=t and b2≠b3.And a1=a4=s≠t=a2=a3.Without loss of generality, we can suppose that s＞t.Let's calculate x-coordinate of intersections. intersection of f1 and f2:(b2－b1)/（s－t）=α intersection of f1 and f3:(b3－b1)/（s－t）=β intersection of f4 and f2:(b2－b4)/（s－t）=γ intersection of f4 and f3:(b3－b4)/（s－t）=δ Without loss of generality, we can suppose that b1＞b4 and b2＞b3.Then β＜α,δ＜γ and α＋δ=(b2＋b3－b1－b4)/（s－t）=β＋γ. Q.E.D.