第28届IMO的第四题是一道关于函数方程的试题：求证不存在函数f：N→N，使得对于每个n∈N，f[f（n）]＝n＋1987[1]．首先，我们把上述试题推广到一般的情形定理1设m为自然数，存在函数f：N→N，使得对每个n∈N，均有f[f（n）]＝n＋m的充要条件是m为偶数证明当。为偶数时，取函 The fourth question of the 28th IMO is a question about functional equations: Prove that there is no function f:N→N, so that for every n∈N, f[f(n)]=n+1987[1]. First, we generalize the above questions to the general case. Theorem 1 Let m be a natural number, and there exists a function f:N→N so that for every n∈N, the necessary and sufficient condition for f[f(n)]=n+m is m is even proof when. When an even number, get a letter