yfqianbj
幼苗
共回答了22个问题采纳率:86.4% 举报
n^k-1=(n^(k-1)+n^(k-2)……+1)(n-1) (a);
(n-1)^2|n^k-1等价于n-1|n^(k-1)+n^(k-2)……+1;
若k被(n-1)整除,则n^(k-1)+n^(k-2)……+1-k=[n^(k-1)-1]+[n^(k-2)-1]……+[1-1];仿造(a),可知每一[]中项均可被n-1整除,从而n^(k-1)+n^(k-2)……+1-k可被n-1整除,进而n-1|n^(k-1)+n^(k-2)……+1;
若n-1|n^(k-1)+n^(k-2)……+1,同样有n^(k-1)+n^(k-2)……+1-k=[n^(k-1)-1]+[n^(k-2)-1]……+[1-1];仿造(a),可知每一[]中项均可被n-1整除,从而n^(k-1)+n^(k-2)……+1-k可被n-1整除,由n-1|n^(k-1)+n^(k-2)……+1有n-1|k;证毕
1年前
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