Thread: 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?

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

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 : )

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

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: , if is odd.
And so this is a multiple of thus it reduces to mod .
If is even then it reduces to .

Try a harder problem: find where is positive integer.

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?

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, as a primitive root and so .

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