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

[teddyayalew]

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)

 

[Mark44]

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)

I get different answers depending on whether p is even or odd.

For example, if p = 6, 1 + 2 + 3 + 4 + 5 $\equiv$ 3 (mod 6).

If p = 5, 1 + 2 + 3 + 4 $\equiv$ 0 (mod 5).

 

[Mark44]

BTW, what you're calculating is (1 + 2 + ... + (p -1))(mod p).

 

[teddyayalew]

Mark, I am sorry I was not clear, when I say p I mean a prime number.

 

[Mark44]

How about if p = 2, the only even prime?

1 ≠ 0 (mod 2)

 

[Robert1986]

Well, you have the right idea by writting
1+2+\cdots+(p-1) = \frac{(p-1)p}{2}

But then you sort of take the long route to the eventual answer.

When you have this:
1 + \cdots + (p-1) = \frac{(p-1)p}{2}

just note that p-1 is even so that (p-1)/2 is an integer. Thus, the sum is divisible by p and so the answer is 0.

But, yes, your answer is correct.

EDIT:
As Mark mentioned, you'll have to do a special case for 2. In fact, this happens a lot in Number Theory.

 

[teddyayalew]

I see.
So if p=2 then I know the sum is : 1 + 1 = 2, which is congruent to 0 mod 2 as well.

Thank you both for you help!

 

[Mark44]

I see.
So if p=2 then I know the sum is : 1 + 1 = 2, which is congruent to 0 mod 2 as well.

Thank you both for you help!

No. If p = 2, the sum has only one term (you quit at 2 - 1 = 1).