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求所有正整数对(a,b)使ab-a²+b+1整除ab+1
人气:492 ℃ 时间:2020-05-11 19:27:22
解答
(ab-a²+b+1)|(ab+1)(ab+1)-(ab-a^2+b+1) = a^2 - b如果a^2 - b = 0那么对于任意的(t,t^2)给出全部ab-a^2+b+1 = ab+1的解否则设b = a + kab-a^2+b+1=a^2+ak-a^2+a+k+1=ak+a+k+1=(a+1)(k+1)ab+1 = a(a+k)+1 = a^2...
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