Thread: powers modulo p and primitive root question

1. powers modulo p and primitive root question

Let p be a prime number.
What is the value of 1 + 2 + 3 +......+(p-1) (mod p)

How would I begin to work this problem?

2. Originally Posted by duggaboy
Let p be a prime number.
What is the value of 1 + 2 + 3 +......+(p-1) (mod p)

How would I begin to work this problem?
Note that (p-k) = -k mod p, k = 1,2,3,4,...

1 + 2 + 3 +......+(p-1) (mod p) = 1 + (p-1)+ 2 + (p-2) +......+(p-1)/2 + (p+1)/2 (mod p) = p + p +......+p (mod p) = 0

3. ok is the idea here that p + p +...+p (modp) =0 because the modulo and the prime number are the same its always eqaul to zero? I'm a little confused why we want it to equal 0....

Thank you so very much : )

4. Originally Posted by duggaboy
ok is the idea here that p + p +...+p (modp) =0 because the modulo and the prime number are the same its always eqaul to zero? I'm a little confused why we want it to equal 0....

Thank you so very much : )
Ya you are right!

p mod p = 0 is the reason

5. Originally Posted by duggaboy
Let p be a prime number.
What is the value of 1 + 2 + 3 +......+(p-1) (mod p)

How would I begin to work this problem?
There is another way: $1+2+...+(p-1) = p\cdot \tfrac{p-1}{2}$, if $p$ is odd.
And so this is a multiple of $p$ thus it reduces to $0$ mod $p$.
If $p$ is even then it reduces to $1$.

Try a harder problem: find $1^k+2^k+...+(p-1)^k (\bmod p)$ where $k$ is positive integer.

6. ok....so if k =2 then we it would be....

1^2 + 2^2 + 3^2 +....+(p-1)^2

(p-1)(p+1)/2p ??

Or would you have to use another approach?? Ultimately you still want 0 mod p right?

7. Originally Posted by duggaboy
ok....so if k =2 then we it would be....

1^2 + 2^2 + 3^2 +....+(p-1)^2

(p-1)(p+1)/2p ??

Or would you have to use another approach?? Ultimately you still want 0 mod p right?
I am not exactly sure what you are doing.
Here is a hint, $p$ as a primitive root $a$ and so $\{ 1,2,...,p-1\} = \{ a,a^2,...,a^{p-1} \}$.

8. n n n
phi(p) = (p - 1) * p 1 * (p - 1) * p 2 ... (p - 1) * p k
1 1 2 2 k k

phi(p) = p-1.

Somthing like this? to get 0 mod p
how would you handle this one? I am very interested in findining out