# Congruence Solving

1. Jun 8, 2011

### Bhatia

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.

2. Jun 8, 2011

### micromass

Staff Emeritus
Hi Bhatia!

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?)

3. Jun 8, 2011

### Bhatia

Thanks for your reply

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.

4. Jun 8, 2011

### micromass

Staff Emeritus
You're working (mod 5) here, so I suggest that you express the x also in (mod 5). (thus x can be 0,1,2,3 or 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).

This is certainly correct, but the above example suggest exactly d!

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

$$a^{\frac{p-1}{d}}$$

By the way, do you know some group theory? It would make it a lot easier...

5. Jun 8, 2011

### Bhatia

Thank you so much...

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 !

Last edited: Jun 8, 2011
6. Jun 8, 2011

### Yuqing

9 is not prime.

7. Jun 8, 2011

### Bhatia

Thanks, Yuqing.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?