|Jun8-11, 12:22 PM||#1|
Let p be a prime number and d / p-1 .
Then which of the following statements about the congruence?
x ^d = 1( mod p) is / are correct :
1. It does not have a solution
2 atmost d incongurent solutions
3 exactly d incongruent solutions
4 aleast d incongruent solutions.
|Jun8-11, 12:36 PM||#2|
What do you think? And why?
(maybe try to work it out for a concrete example of p and d? Like p=11 and d=5?)
|Jun8-11, 01:05 PM||#3|
Going by your idea
Let p= 5, then since d / p-1 then I guess we have d= 1, 2, 4...then solving x^ d = 1 (mod 5)
for d=1, x=1
for d=2, x= 4, 6,9,11,14, ....
for d=4, x= 2, 3,4,6, 7 .....
So I am thinking atleast d....
But not sure....I did not study number theory at graduation level.
|Jun8-11, 01:26 PM||#4|
When I do this, I get
for d=1: x=1
for d=2: x=1,4
for d=4: x=1,2,3,4
Solutions like 9 are superfluous here, since 9=4 (mod 5).
To prove this consider a number a coprime with p (for example take a=2 when p is odd). Then what can you say about
By the way, do you know some group theory? It would make it a lot easier...
|Jun8-11, 03:58 PM||#5|
I agree with this...I realized my mistake x should be strictly less than p.
Taking p to be odd (as suggested) p= 5 , then d =1, 2, 3, 4.
Does 2 ^ (p-1)/d imply the following we get:
for d=1, 2^ 4 = 16 works for d=1 that is , 16= 1 (mod 5).
for d=2, 2 ^ 2= 4 works for d=2 that is, 4^2 = 1( mod 5)
for d= 4, 2 ^ 1 =2 works for d =4, that is , 2^ 4 = 1 (mod 5).
Does this mean that we can generate other values of x ( above we get x= 16)....so x has to be at least d but can be more than d as well.
Please help !
|Jun8-11, 04:01 PM||#6|
9 is not prime.
|Jun8-11, 04:04 PM||#7|
|Similar Threads for: Congruence Solving|
|solutions to na=0 (mod m)||Linear & Abstract Algebra||4|
|Solving a linear congruence||Calculus & Beyond Homework||0|
|Congruence - subject difficulty||General Math||0|
|Solving a congruence||Calculus & Beyond Homework||2|
|congruence help||Linear & Abstract Algebra||5|