1. ## Complex prime problem

I am having great difficulty figuring the followin problem. Your help is appreciated!

Find all primes p such that 2011p = 2+3+4+...+n for some positive integer n

I've goten as far as 2011p = (n(n+1)/2)-1

p= ((n(n+1)/2)-1)/2011 and plugging in various numbers for n.

Am i on the right track?

2. Originally Posted by fernipascual
I am having great difficulty figuring the followin problem. Your help is appreciated!

Find all primes p such that 2011p = 2+3+4+...+n for some positive integer n

I've goten as far as 2011p = (n(n+1)/2)-1

p= ((n(n+1)/2)-1)/2011 and plugging in various numbers for n.

Am i on the right track?

I think you are. Observe that $\displaystyle{2,011p=\frac{n(n+1)}{2}-1=\frac{(n+2)(n-1)}{2}}$ ...

The above already tells you that if $p>2$ then either $n=0\!\!\pmod 4\mbox{ or else } n=3\!\!\pmod 4$ .

Tonio

3. Thanks much Tonio. What does mod 4 mean? Is n=3 the only answer?