证(x-y)^5+(y-z)^5+(z-x)^5能被5(y-z)(z-x)(x-y)整除(xyz是整数且两两不等)

Question

证(x-y)^5+(y-z)^5+(z-x)^5能被5(y-z)(z-x)(x-y)整除(xyz是整数且两两不等)

Answer 1

令x-y=a,y-z=b 原式=a^5+b^5-(a+b)^5展开系数都能被5整除,所以能被5整除.
因为a^5+b^5=(a+b)(a^4-a^3b+a^2b^2+ab^3+b^4)
故原式=(x-y+y-z)(…)+(z-x)^5 能被(z-x)整除
同理能被(y-z)和(x-y)整除
综上：能被5(y-z)(z-x)(x-y)整除.