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ax - b l f(x) => a - b l f(1)

Jul 6th 2013, 07:58

a.b為整數且互質當ax - b l f(x) => a - b l f(1) , - a - b l f(-1) 且對任一整數m而言 , 則必 ma - b l f(m) , 故知如果存在一整數 m , 使 ma - b 不l f(m) , 則可斷言 ax -b 不l f(x).

上面是對.

設 f(x) 為整係數 , 若 3a -b 不l f(3) , 則 ax - b 不l f(x)

這句話錯了, 我當然知道他錯. 只是錯在哪說明一下就好.

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