Here we introduce a simple elementary method due to myself. Let n be nk(p) or nk(p) − 1 according to whether −1 is a kth power residue mod p. Note that −n is a kth power nonresidue mod p. For i = 1, · · · , nk(p) − 1, clearly p − in is a kth power nonresidue mod p; if p − in > 0 then we must have p− in > nk(p) > n and hence p− (i+ 1)n > 0 since p is a prime. As p− n > 0, by the above p− nk(p)n ...