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?

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- Jun 14th 2008, 07:45 AMduggaboypowers 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? - Jun 14th 2008, 08:26 AMIsomorphism
- Jun 14th 2008, 08:57 AMduggaboy
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 : ) - Jun 14th 2008, 09:00 AMIsomorphism
- Jun 14th 2008, 06:02 PMThePerfectHacker
There is another way: $\displaystyle 1+2+...+(p-1) = p\cdot \tfrac{p-1}{2} $, if $\displaystyle p$ is odd.

And so this is a multiple of $\displaystyle p$ thus it reduces to $\displaystyle 0$ mod $\displaystyle p$.

If $\displaystyle p$ is even then it reduces to $\displaystyle 1$.

Try a harder problem: find $\displaystyle 1^k+2^k+...+(p-1)^k (\bmod p)$ where $\displaystyle k$ is positive integer. - Jun 16th 2008, 09:57 AMduggaboy
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? - Jun 16th 2008, 10:13 AMThePerfectHacker
- Jun 16th 2008, 10:57 AMduggaboy
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