推 kevin092178 :人帥真好 05/25 15:41
※ 引述《j19951102 (j19951102)》之銘言:
: 已知n為正整數,p為質數,且滿足條件n|(p-1)與p|(n^3-1),
: 試證:4p-3必為某整數的完全平方。
: 謝謝!
寫寫另一種方法:D
p|(n^3-1)
=> p|n-1 (-><-) or p|n^2+n+1
write ap = n^2+n+1 , where a in N
=> a(p-1)= n^2+n+(1-a)
=> p-1|n^2+n+(1-a)
∵ n|p-1
∴ n|n^2+n+(1-a)
=> n|1-a
=> n|a-1
if a≠1 ,
then n≦a-1 => a≧n+1
and we have n|p-1
=> n≦p-1 => p≧n+1
=> (n+1)^2 ≦ ap = n^2+n+1 (-><-)
so a = 1
then p = n^2+n+1
=> 4p-3 = 4n^2+4n+1 = (2n+1)^2 .
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