Forgot password?
 Create new account
View 113|Reply 1

[数论] $x^2+y^2\equiv 1\pmod p$的解的数量

[Copy link]

3147

Threads

8497

Posts

610K

Credits

Credits
66183
QQ

Show all posts

hbghlyj Posted at 2024-11-14 18:36:08 |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
$$\begin{cases}(p+3)/4&\text{ if } p\equiv 1\pmod 4\\(p+5)/4&\text{ if
}p\equiv 3\pmod 4\end{cases}$$

3147

Threads

8497

Posts

610K

Credits

Credits
66183
QQ

Show all posts

 Author| hbghlyj Posted at 2025-1-13 17:10:19
顶一下

手机版Mobile version|Leisure Math Forum

2025-4-21 01:29 GMT+8

Powered by Discuz!

× Quick Reply To Top Return to the list