很偶然的, 看到了几个韩京俊传出来的数论问题. 据说问题来自牟晓生.

1. 设 \(p\) 为大于 \(3\) 的素数, 证明 \(\dfrac{p^p-1}{p-1}\) 和 \(\dfrac{p^p+1}{p+1}\) 不能都是素数幂;
2. 设 \(n\gt5\), 证明 \(n!\) 不能整除它的正约数之和;
3. 设 \(A\), \(B\) 划分正整数集, 如果\(A+A\) 和 \(B+B\) 都只含有有限个素数, 证明\(A\) 或 \(B\) 是全体奇数的集合;
4. 设 \(M\) 是给定正整数, 证明对每个充分大的素数 \(p\), 存在\(M\)个连续的 \(\bmod p\) 的二次非剩余;
5. 设 \(q\) 是一个不大于\(\dfrac{\pi^2}6 -1\) 的正有理数, 证明 \(q\) 可写为若干互异单位分数的平方和;
6. 对每个充分大的正整数 \(k\), 存在若干互异正整数, 其和为 \(k\), 其倒数和为 \(1\);
7. 在 \(n^2\) 和 \((n+1)^2\) 间总有一些正整数的积是一个平方数的两倍;
8. 若一些单位根之和在单位圆上, 则必亦为单位根;
9. 设 \(f(x)=a_0+a_1x+a_2x^2+\dotsb\) 是一个整系数的形式幂级数, 假定 \(\dfrac{f^\prime(x)}{f(x)}\) 也是一个整系数的形式幂级数, 证明对任意下标 \(k\), \(a_k\) 能被 \(a_0\) 整除.

这些问题显然非常的有难度. 第 3 个问题, 俺多年前就见过, 是 Paul Erdős 在美国数学月刊上提的问题(编号 A6431).

俺特意调查了其他几个问题的出处.

问题 5 也是 Paul Erdős 提出的, 但证明是 R.L. Graham (也可能是 Sierpsinski) 给的. R.L. Graham 证明了

\(\dfrac pq\) can expressed as the finite sum of reciprocals of distinct squares if and only if

\[\frac pq\in[0, \frac{\pi^2}6-1)\cup[1,\frac{\pi^2}6).\]

问题 6 的答案也是 R. L. Graham 提供的: Graham published a proof  in 1963 as “A Theorem on Partitions”, Journal of the Australian Mathematical Society 3 (1963), pp. 435-441.

If  \(n\) is an integer exceeding \(77\) then there exist positive integers \(k\), \(a_1\), \(a_2\), \(\dotsc\), \(a_k\) such that:

1. \(1\lt a_1\lt a_2\lt \dotsc \lt a_k;\)
2.  \(a_1+ a_2+ \dotsb + a_k=n;\)
3.  \(\frac1{a_1}+ \frac1{a_2}+ \dotsb + \frac1{a_k}=1.\)

His proof is constructive and fairly short, but it does require a long table of decompositions for relatively small values of \(n\). It would be interesting to see a non-constructive proof that doesn’t require such a long list.

问题 7 也不简单.

Granville and Selfridge, Product of integers in an interval, modulo squares: “We prove a conjecture of Irving Kaplansky which asserts that between any pair of consecutive positive squares there is a set of distinct integers whose product is twice a square.”

The details are Electronic Journal of Combinatorics, Volume 8(1), 2001.

有比问题 8 更普遍的结果. More precisely, let \(\zeta_1\), \(\dotsc\), \(\zeta_k\) be \(n\)-th roots of unity. If

\[|\sum_{i=1}^k n_i\zeta_i|= 1,\]

where \(n_i\in\mathbb Z\), then \(\sum\limits_{i=1}^k n_i \zeta_i\) is also an \(n\)-th root of unit.