設(shè)
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)
我們考察f(x)'的次數(shù),是n-1次,而且f(x)'的n-1次項(xiàng)的系數(shù)是n,所以可以設(shè)
(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等于某個(gè)a[i]
不妨設(shè)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])
左邊每一項(xiàng)都能被(x-a[1])整除,所以右邊(x-a[2]),...,(x-a[n])必然有一項(xiàng)滿足a[j]=a[1]
不妨設(shè)是a[2]=a[1],那么利用a[1]=m我們有：
[(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])
現(xiàn)在左邊每項(xiàng)可以被x-m整除,所有右邊必然有一項(xiàng),可以被x-m整除,不妨設(shè)a[3]=m.
然后.仿照上面的步驟,我們得到：
a[i]都是m
完畢.