On the addition of residue classes mod p
On the addition of residue classes mod p
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DOI：
10.4064/aa-9-2-149-159
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发表时间：
1964
期刊：
影响因子：
0.7
通讯作者：
P. Erdös;H. Heilbronn
中科院分区：
文献类型：
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作者：
P. Erdös;H. Heilbronn
在本文中，我们研究以下问题。设\(p\)为素数，\(\alpha_1,\cdots,\alpha_k\)是模\(p\)的不同非零剩余类，\(N\)是模\(p\)的一个剩余类。用\(D\)表示同余式\(e_1\alpha_1 + \cdots + e_k\alpha_k\equiv N(\bmod p)\)的解的个数，其中\(e_1,\cdots,e_k\)限于\(0\)和\(1\)。关于函数\(D(N)\)能说些什么呢？我们证明两个定理。
定理\(I\)。如果\(k > 3(\log p)^{\frac{1}{2}}\)，则\(D(N)>0\)。
定理\(II\)。当\(p\rightarrow\infty\)时，\(D(N)=2^kp^{-1}(1 + o(1))\)，如果\(k^3p^{\frac{1}{2}}\rightarrow\infty\)。
定理\(I\)几乎是最优的。令\(\alpha_1 = 1\)，\(\alpha_2=-1\)，\(\alpha_3 = 2\)，\(\alpha_4=-2\)，\(\cdots\)，\(\alpha_k = (\frac{1}{2})^{k - 1}[\frac{1}{2}(k + 1)]\)。那么通过一个简单计算可得，如果\(k\geq1\)，当\(p\rightarrow\infty\)时\(D(\frac{1}{2}(p - 1)) = 0\)。
在证明方法上，这两个定理有很大不同。定理\(I\)的证明是初等的，完全依赖于模\(p\)剩余类的运算，而定理\(II\)的证明基于有限傅里叶级数的应用以及对丢番图逼近的简单考虑。在附录中，我们陈述了各种我们无法证明的进一步猜想。
In this paper we investigate the following question. Let p be a prime, a,, " ', cck distinct non-zero residue classes modp, N a residue class modp. denote Dhe number of solutions of the congruence ela,+... + ekak = N(modp) where the e,,. .. , ek are restricted to the values 0 and 1, What can be said about the function J?(N)? We prove two theorems. THEOREM I. 14 " (X) > 0 if k > 3 (6;~) " ~. THEOREM II. P(N) = 2kp-'(1+o(1)) $ k3pS2-+ 00 as p + co. Theorem I is almost best possible. Put al = ', '2 =-1, a3 = 2, a4 =-2,. .. . ckk = (Lx)k-'[&(k+l)]. Then it follows from an easy calc.ulation that F(S(p-1)) = 0 if k 1. p-too In the method of proof the two theorems differ considerably. The proof of Theorem I is elementary, depending entirely on the manipulation of residue classes mBdp, whereas the proof of Theorem II is based on the application of finite Fourier series and simple considerations on diophantine approximations. In an appendix we state various further conjectures which we are not able to prove.