关于同余式sum from i=1 to k(x_i^2≡0(mod p))(1≤x_1<x_2<…<x_k≤(P-1)/2)的解数公式

On the Formula to Calculate the Number of Solutions of sum from i=1 to (1/i)x^2=0(modp)

设N_K为同余方程x_1~2+…+x_k^2≡0(modp),1≤x_1<x_2<…<x_k≤P-1/2的解(x_1,…,x_k)的个数,这里p是一个奇素数.本文给出了N_4的计算公式以及求N_k的计算公式的一个算法.

Let p be an odd prime and N1 denote the numbers of solutions(x1,…,xk) of congruent equation In this paper, we obtain a formula to calculate N4. We also give an algorithm for obtaining a formula to calculate Nk.