MathLabs

第4题

确定所有 pairs (a,b)(a,b) 的正整数 使得 ab2+b+7 ab^{2}+b+7 译文: divides a2b+a+b a^{2}b+a+b .
第 2/6 步:Handle zero remainder
通俗地说

译文:The zero remainder gives an infinite family.

7a=b2⇒(a,b)=(7t2,7t)7a=b^2\Rightarrow (a,b)=(7t^2,7t)
详细分析

若 7a=b27a=b^2 译文:, then 7∣b7\mid b 译文:, say b=7t b=7t , 且consequently a=7t2 a=7t^2. 反之, 对于these values E=tD E=tD , so 每个 (7t2,7t)(7t^2,7t) 是a 解.