设x、y、z为整数,证明:x^4*(y-z)+y^4*(z-x)+z^4*(x-y)/(y+z)^2+(z+x)^2+(

x^4(y-z)+y^4(z-x)+z^4(x-y)
=xy(x^3-y^3)+yz(y^3-z^3)+zx(z^3-x^3)
=xy(x^3-y^3)+yz(y^3-z^3)-zx[(x^3-y^3)+(y^3-z^3)]
=x(y-z)(x^3-y^3)+z(y-x)(y^3-z^3)
=(x-y)(y-z)[x(x^2+xy+y^2)-z(y^2+yz+z^2)]
=(x-y)(y-z)[(x^3-z^3)+(x^2-z^2)y+(x-z)y^2]
=(x-y)(y-z)(x-z)(x^2+y^2+z^2+xy+yz+zx)
又(x+y)^2+(y+z)^2+(z+x)^2=2(x^2+y^2+z^2+xy+yz+zx)
所以[x^4(y-z)+y^4(z-x)+z^4(x-y)]/[(x+y)^2+(y+z)^2+(z+x)^2]
=(x-y)(y-z)(x-z)/2
又因为x-y,y-z,x-z三个数中至少有一个是偶数
所以(x-y)(y-z)(x-z)/2为整数,证毕