关于幂残模素数的几个注意事项

D. Mej'ia, Y. Kiriu
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引用次数: 0

摘要

设q为质数。我们对奇数素数p≠q进行分类,使得方程x2≡q (mod p)有解,具体地说,我们找到具有4q(模4q)的整数相对素数的乘法群U4q的子群L4q,使得x2≡q (mod p)对某c∈L4q有解iff p≡c (mod 4q)。并且,L4q是U4q的半阶子群中唯一包含−1的子群。考虑环Z[√2],对于任何奇素数p,已知方程x2≡2 (mod p)有解,只要方程x2−2y2 = p在整数中有解。我们问这是否可以在n≥2的Z[n√2]的情况下推广,即:对于任意素数p≡1 (mod n),是否xn≡2 (mod p)在方程D2n(x0,…)下有解。, xn−1)= p在整数中有解?这里D2n(x ā)表示Q的域扩展Q(n√2)的范数。我们解决了这个问题的一些弱版本,其中与p的等式被0 (mod p)(可被p整除)取代,并且考虑对任意r∈Z代替2的“范数”Drn(x ā)。
本文章由计算机程序翻译,如有差异,请以英文原文为准。
Some notes about power residues modulo prime
Let q be a prime. We classify the odd primes p ≠ q such that the equation x2 ≡ q (mod p) has a solution, concretely, we find a subgroup L4q of the multiplicative group U4q of integers relatively prime with 4q (modulo 4q) such that x2 ≡ q (mod p) has a solution iff p ≡ c (mod 4q) for some c ∈ L4q. Moreover, L4q is the only subgroup of U4q of half order containing −1. Considering the ring Z[√2], for any odd prime p it is known that the equation x2 ≡ 2 (mod p) has a solution iff the equation x2 −2y2 = p has a solution in the integers. We ask whether this can be extended in the context of Z[n√2] with n ≥2, namely: for any prime p ≡ 1 (mod n), is it true that xn ≡ 2 (mod p) has a solution iff the equation D2n(x0, . . . , xn−1) = p has a solution in the integers? Here D2n(x̄) represents the norm of the field extension Q(n√2) of Q. We solve some weak versions of this problem, where equality with p is replaced by 0 (mod p) (divisible by p), and the “norm" Drn(x̄) is considered for any r ∈ Z in the place of 2.
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