Recent content by reyomit

I was under the impression that \Re^{+} denoted the positive reals and not the non-negative reals.

Forgive me if I am missing something, but is there anything mapping to (0,0,0)?

I feel much the fool. Thank you.

Let me be more explicit, using that (m-a,a)=1 whenever (m,a)=1, I concluded (m-r1)+(m-r2)+...+(m-rx)=r1+...+rx mod m and also (m-r1)+(m-r2)+...+(m-rx)= -r1-r2-...-rx mod m. So -(r1+...+rx)=r1+...+rx mod m or 2(r1+...+rx)=0 mod m.

I see that (a,m)=1 immediately implies that (m-a,m)=1,and used that to prove that 2(r1 + r2 + ... +rx) = 0 mod m. Then when m is odd 2 has a multiplicative inverse mod m, and we can just multiply by that to get r1 + r2 + ... +rx = 0 mod m. But when m is even, I'm not totally sure, what...

Homework Statement Prove that if {r1,r2,...rx} is a reduced residue system mod m (where x=\phi(m), m>2), then r1 + r2 + ... + rx= 0 mod m. Homework Equations The Attempt at a Solution I've been able to prove it pretty simply for odd m and for m=2k where k is odd, but for m with higher powers...