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## Homework Statement

What is the value of 1 + 2 + 3 +...+ (p-1)(mod p)?

## Homework Equations

p = 0 (mod p)

p-1 = -1 (mod p)

1 + 2 + 3 + ...+n = n(n+1)/2

## The Attempt at a Solution

I know 1 + 2 + 3 +...+ (p-1) = (p-1)(p)/2

I worked the problem, but i don't know if i am correct:

work: i am looking for a b s.t.

b = (p-1)(p)/2 (mod p) or 2b = (p-1)(p) ( mod p)

we know (p-1) = -1 (mod p)

so : (p-1)p = -p (mod p)

this implies : 2b = -p(mod p) ,also we know p = 0 (mod p) so b =0 (mod p)

So the answer is 1 + 2 + 3 +...+ (p-1) = 0 (mod p)