A math contest: X? +y？ +u？ +v？ = 4 Xu+yv+XV+Yu = 0 xyu+YUV+uvx+vxy =-2 xyvu =- 1

Solution: ∫ Xu+YV+XV+Yu = X (U+V)+Y (V+U) = (X+Y) (V+U) = 0.

That is, ∴x+y=0 or v+u=0.

X=-y or v =-u.

∵(x+y+v+u)？

=[(x+y)+(v+u)]？

=(x+y)？ +(v+u)？ +2(x+y)(v+u)

=x？ +y？ +2xy+v？ +u？ +2vu+2(XV+yv+ Xu+Yu)

=x？ +y？ +v？ +u？ +2xy+2vu+2(XV+yv+ Xu+Yu)

=4+2xy+2vu

{xu+yv+xv+yu=0}

Also: xyu+yuv+uvx+vxy.

=xy(u+v)+uv(x+y)

=-2

1) When

When x+y=0, x =-y.

(x+y+v+u)？ =(v+u)？ =4+2xy+2vu=4-2x？ +2vu, that is, (v+u)? =4-2x？ +2vu

∫xyvu =- 1，x==-y

∴-x？ Vu=- 1, which is x? = 1/vu

∫-2 = XYU+YUV+UVX+VXY = XY(U+V)+UV(X+Y)=-X？ (u+v), that is, x? =2/(u+v)

∴ 1/vu=2/(u+v), that is, 2vu = u+v.

Will it be X? = 1/vu, 2vu=u+v replacement (v+u)? =4-2x？ +2vu

(2vu)？ =4-2/vu+2vu, so t=vu is (2t)? =4-2/t+2t

4(t？ - 1)=2(t- 1/t)

2(t？ - 1)=(t？ - 1)/t

answer

What time? -1=0, namely (vu)? = 1, vu= 1 (due to x? = 1/vu, so give up the negative root)

∫u+v = 2vu, ∴u+v=2, ∴u=v= 1 and (u+v)? =4=v？ +u？ +2vu=v？ +u？ +2, which is v? +u？ =2

∴x？ +y？ =2

∫x =-y, ∴x= 1 and y=- 1 or x=- 1 and y= 1.

B) when t? - 1≠0

T= 1/2, that is, vu= 1/2, then u+v=2vu= 1.

∫u+v = 1，vu= 1/2

∴v= 1/2u, that is, u+ 1/2u= 1.

2u？ -2u+ 1=0

{At this time, the equation has no solution, so give up}

2) when v+u=0

In the same way.

A: These four numbers are 1,-1, 1, 1 respectively.