论文标题
有关电源残留的一些笔记Modulo Prime
Some notes about power residues modulo prime
论文作者
论文摘要
令$ Q $为素数。我们对奇数$ p \ neq q $进行了分类，使得方程$ x^2 \ equiv q \ pmod {p} $具有解决方案，具体地说，我们找到了一个子组$ \ mathbb {l} _ {4q} $ \ sathbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb {u} $ priend $ { $ 4Q $）使得$ x^2 \ equiv q \ pmod {p} $具有解决方案，即iff $ p \ equiv c \ equiv c \ pmod {4q} $，对于某些$ c \ in \ in \ mathbb {l} _ {4q} $。此外，$ \ mathbb {l} _ {4Q} $是$ \ Mathbb {u} _ {4Q} $的唯一子组，其中包含$ -1 $。 考虑到环$ \ mathbb {z} [\ sqrt {2}] $，对于任何奇数$ p $，众所周知，公式$ x^2 \ equiv 2 \ equiv 2 \ pmod {p} $具有解决方案，如果方程式$ x^2-2y^2 = p $在整数中都有解决方案。我们询问是否可以在$ \ mathbb {z} [\ sqrt [n] {2}] $的情况下使用$ n \ geq 2 $，即$ d^2_n（x_0，\ ldots，x_ {n-1}）= p $在整数中有一个解决方案吗？这里$ d^2_n（\ bar {x}）$代表$ \ mathbb {q} $的字段扩展$ \ mathbb {q}（\ sqrt [n] {2}）$的标准。我们解决了此问题的一些薄弱版本，其中$ p $的平等被$ 0 \ pmod {p} $（由$ p $除外），而“ norm norm” $ d^r_n（\ bar {x}）$被考虑在\ mathbb {z} $ 2 $ 2 $ 2 $的情况下考虑任何$ r \ in \ mathbb {z} $。
Let $q$ be a prime. We classify the odd primes $p\neq q$ such that the equation $x^2\equiv q\pmod{p}$ has a solution, concretely, we find a subgroup $\mathbb{L}_{4q}$ of the multiplicative group $\mathbb{U}_{4q}$ of integers relatively prime with $4q$ (modulo $4q$) such that $x^2\equiv q\pmod{p}$ has a solution iff $p\equiv c\pmod{4q}$ for some $c\in\mathbb{L}_{4q}$. Moreover, $\mathbb{L}_{4q}$ is the only subgroup of $\mathbb{U}_{4q}$ of half order containing $-1$. Considering the ring $\mathbb{Z}[\sqrt{2}]$, for any odd prime $p$ it is known that the equation $x^2\equiv 2\pmod{p}$ has a solution iff the equation $x^2-2y^2=p$ has a solution in the integers. We ask whether this can be extended in the context of $\mathbb{Z}[\sqrt[n]{2}]$ with $n\geq 2$, namely: for any prime $p\equiv 1\pmod{n}$, is it true that $x^n\equiv 2\pmod{p}$ has a solution iff the equation $D^2_n(x_0,\ldots,x_{n-1})=p$ has a solution in the integers? Here $D^2_n(\bar{x})$ represents the norm of the field extension $\mathbb{Q}(\sqrt[n]{2})$ of $\mathbb{Q}$. We solve some weak versions of this problem, where equality with $p$ is replaced by $0\pmod{p}$ (divisible by $p$), and the "norm" $D^r_n(\bar{x})$ is considered for any $r\in\mathbb{Z}$ in the place of $2$.
