Forgot password?
 Register account
View 231|Reply 1

[数论] $x^2+y^2\equiv 1\pmod p$的解的数量

[Copy link]

3153

Threads

7905

Posts

610K

Credits

Credits
64091
QQ

Show all posts

hbghlyj Posted 2024-11-14 18:36 |Read mode
A Classical Introduction to Modern Number Theory by Ireland and Rosen
对任意奇素数 $p$,集合 $$A_p=\{(x,y):0\le x\le(p-1)/2,\, 0\le y\le(p-1)/2,\, x^2+y^2\equiv 1\pmod p\}$$ 中的元素数量为
$$\begin{cases}(p+3)/4&\text{ if } p\equiv 1\pmod 4\\(p+5)/4&\text{ if
}p\equiv 3\pmod 4\end{cases}$$

3153

Threads

7905

Posts

610K

Credits

Credits
64091
QQ

Show all posts

 Author| hbghlyj Posted 2025-1-13 17:10
顶一下

Mobile version|Discuz Math Forum

2025-6-5 07:43 GMT+8

Powered by Discuz!

× Quick Reply To Top Edit