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[数论] 求解

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大佬最帅 Posted 2021-12-23 21:01 |Read mode
Last edited by 大佬最帅 2021-12-23 21:38问题:能否找到2021个互不相同的正整数,使得任意两数的和能被他们的差整除
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爪机专用 Posted 2021-12-23 21:13
回复 1# 大佬最帅

图片里就一行字,又没公式,请去图,码字!

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 Author| 大佬最帅 Posted 2021-12-23 21:39
回复 2# 爪机专用
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realnumber Posted 2021-12-24 16:16
Last edited by realnumber 2021-12-24 16:28若单调递增正数列{$a_n$}符合每两项的和能被他们的差整除,
记$s=\prod_{1\le i<j\le n}(a_j-a_i)$那么有这样性质
1.数列{$s+a_n$}也符合要求.2.最前面可以添一项0.
这样可以构造出2021项,
比如:1,2,3符合3项要求,0,1,2,3,每一项加$s_1$=12
那么12,13,14,15也符合,0,12,13,14,15,每一项加$s_2$=$12\times13\times14\times15\times s_1$,也符合.

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