∵n|p-1,∴可设p-1=kn(k为正整数)≥n,∴p≥n+1＞n＞n-1∵p|n³-1,∴p|(n-1)(n²+n+1),注意到p为素数,且p＞n-1,∴p与n-1互素,∴p|n²+n+1,∴p|n²+n+1-p,即p|n²+n-kn,注意到p也与n互素,∴p|n-k+...