To prove this problem, it is actually to find a qualified f(x) so that for any x belonging to [0. 1], there is f(x)≠f(x+l), where l≠ 1/n (in fact, l = 1 Therefore, a continuous function g(x) on [0. 1] is constructed, which satisfies g(0)=0, g( 1)=m≠0, g(x+l)=g(x), and then the function f (x) = g (x. So far, we have found a qualified F, which is continuous on [0, 1] and satisfies f(0)=f( 1), but for any x, l≠ 1/n, there is f (x) ≠ f.