|
本帖最后由 青青子衿 于 2024-1-14 14:20 编辑 能不能利用这个三角函数形式的公式推出模7勒让德符号的平方有理形式或单个取整形式?
如下形式只算作枚举形式,不算平方有理形式或单个取整形式
\begin{align*}
\left(\frac{7}{p}\right)=\left\{ \begin{aligned}&1~\text{if}~p \equiv \pm 1, \pm 3, \pm 9 ~(\text{mod}~ 28) \\ -&1~\text{if}~p \equiv \pm 5, \pm 11, \pm 13~(\text{mod}~28) \end{aligned} \right.
\end{align*}
类似于如下形式的
\begin{align*}
\left(\frac{-1}{p}\right) &= (-1)^{\frac{p-1}{2}}\\
\left(\frac{2}{p}\right) &= (-1)^\tfrac{p^2-1}{8}\\ &=(-1)^{\left\lfloor \frac{p+1}{4}\right\rfloor }\\
\left(\frac{3}{p}\right) &= (-1)^{\big\lfloor \frac{p+1}{6}\big\rfloor}\\
&=(-1)^{\left\lfloor \frac{p+1}{5}\right\rfloor }\\
\left(\frac{5}{p}\right) &=(-1)^{\big\lfloor \frac{p+2}{5}\big \rfloor}\\
&=(-1)^{\big\lfloor \frac{2p+2}{5}\big \rfloor}
\end{align*}
- Clear["Global`*"]
- s = 7;
- lst = Select[Range[0, 4 s - 1], GCD[#, 4 s] == 1 &]
- Select[Flatten[
- Table[{a, b, c}, {a, 1, 100}, {b, 1, 100}, {c, 1, 100}],
- 2], (-1)^Floor[(#[[1]]*lst + #[[2]])/#[[3]]] ==
- JacobiSymbol[s, lst] &]
- JacobiSymbol[s, lst]
- (-1)^Floor[(lst + 2)/7]
复制代码 |
|