又有数学题求解，先谢谢了！

1. Find all solution: a^2003 = 1 (mod 17)

2. Find all solutions to each congruence. Reminder: a=b mod m means than m|b-a, equivalently, that band a leave the same remainder when divided by m, so that eg 5=12 = -2 (mod 7).

a) 3x = 1 (mod 11) b. 20x=52 (mod 7) c. 20x = 52 (mod 88)

我一见 mod 就晕了。怎么找所有的解啊！谢谢！

所有跟帖： 

• Did you forget to add some conditions? -slowguy- ♂ (68 bytes) () 04/23/2012 postreply 13:23:17

• there are no more condition in the problems -琼岛春荫- ♀ (44 bytes) () 04/23/2012 postreply 13:44:42

• 1) We shuold use number theory for strict answer. -jinjing- ♀ (564 bytes) () 04/23/2012 postreply 15:47:27

• 你,你，你也太厉害了！ -琼岛春荫- ♀ (76 bytes) () 04/23/2012 postreply 17:25:45

• Not worth it. Just enjoying your life, and leave the problems to -slowguy- ♂ (174 bytes) () 04/23/2012 postreply 19:41:46

• 不敢当, -jinjing- ♀ (0 bytes) () 04/23/2012 postreply 20:31:55

• 这是解线性同余。 -calligraphy- ♂ (515 bytes) () 04/23/2012 postreply 19:39:56

• 我居然看懂了，有个问题 -琼岛春荫- ♀ (68 bytes) () 04/24/2012 postreply 12:15:11

• after the first one 7, you add 22 (88/4) and can get other three -calligraphy- ♂ (0 bytes) () 04/24/2012 postreply 12:51:03

• 问好！ ：） -思如昀- ♀ (0 bytes) () 05/11/2012 postreply 19:30:27

• try to solve the 1st one -calligraphy- ♂ (324 bytes) () 04/23/2012 postreply 20:32:24

• 为什么a^3 = 1(mod 17), a=1? -琼岛春荫- ♀ (101 bytes) () 04/24/2012 postreply 11:31:58

• 试着用二项式定理理解一下。 -calligraphy- ♂ (0 bytes) () 04/24/2012 postreply 12:52:19

• 你的意思是<17的只有 1.其他都是重复的对吗？谢了 -琼岛春荫- ♀ (0 bytes) () 04/24/2012 postreply 11:40:47

• 17 !/ (i^3-1), i=2,3,...,15,16. -jinjing- ♀ (0 bytes) () 04/25/2012 postreply 13:31:35

• 这些不是常规的高中数学吧？ -calligraphy- ♂ (0 bytes) () 04/23/2012 postreply 20:40:39

• 回复：这些不是常规的高中数学吧？ -琼岛春荫- ♀ (21 bytes) () 04/24/2012 postreply 11:32:36

• 谢谢上面各位！ -琼岛春荫- ♀ (71 bytes) () 04/24/2012 postreply 11:01:51

• 几十年前,读过"数论导引",有两条定理,证明较长.现简叙如下: -jinjing- ♀ (265 bytes) () 04/24/2012 postreply 13:41:24

• 您可改2003为2012....(2012,16)=4,...有4组解. -jinjing- ♀ (59 bytes) () 04/24/2012 postreply 14:48:19