设
f(x)=a(x-a[1])(x-a[2])...(x-a[n])
那么
f(x)'=a[(x-a[2])(x-a[3])...(x-a[n])+(x-a[1])(x-a[3])...(x-a[n])+(x-a[1])(x-a[2])(x-a[4])...(x-a[n])+.+(x-a[1])(x-a[2])...(x-a[n-1])]
如果f(x)'|f(x)
我们考察f(x)'的次数,是n-1次,而且f(x)'的n-1次项的系数是n,所以可以设
(x-m)f(x)'=nf(x)
也就是
(x-m)f(x)'=na(x-a[1])(x-a[2])...(x-a[n])
所以必然m等于某个a[i]
不妨设m=a[1]
那么
(x-a[1])a[(x-a[2])(x-a[3])...(x-a[n])+(x-a[1])(x-a[3])...(x-a[n])+(x-a[1])(x-a[2])(x-a[4])...(x-a[n])+.+(x-a[1])(x-a[2])...(x-a[n-1])]=na(x-a[1])(x-a[2])...(x-a[n])
那么
[(x-a[1])(x-a[3])...(x-a[n])+(x-a[1])(x-a[2])(x-a[4])...(x-a[n])+.+(x-a[1])(x-a[2])...(x-a[n-1])]=(n-1)(x-a[2])(x-a[3])...(x-a[n])
左边每一项都能被(x-a[1])整除,所以右边(x-a[2]),...,(x-a[n])必然有一项满足a[j]=a[1]
不妨设是a[2]=a[1],那么利用a[1]=m我们有：
[(x-m)(x-a[3])...(x-a[n])+(x-m)(x-m)(x-a[4])...(x-a[n])+.+(x-m)(x-m)(x-a[3])...(x-a[n-1])]=(n-1)(x-m)(x-a[3])...(x-a[n])
那么
[(x-a[3])...(x-a[n])+(x-m)(x-a[4])...(x-a[n])+.+(x-m)(x-a[3])...(x-a[n-1])]=(n-1)(x-a[3])...(x-a[n])
那么
[(x-m)(x-a[4])...(x-a[n])+.+(x-m)(x-a[3])...(x-a[n-1])]=(n-2)(x-a[3])...(x-a[n])
现在左边每项可以被x-m整除,所有右边必然有一项,可以被x-m整除,不妨设a[3]=m.
然后.仿照上面的步骤,我们得到：
a[i]都是m
完毕.