Question

I suspect this to be true, not sure how I'd go about proving it though.

Answer 1

Um... @agent0smith

Answer 2

Heh, whoops upper index should be this:
\[(b+a-1)(b-a) \equiv \sum_{n=a}^{b-1} 2*n \mod m\]

Answer 3

Welcome back empty

Answer 4

Haha thanks. I think I figured it out actually, I just realized I could prove this one. I think I'll present the real situation later if I get bored enough.