Let n be a positive integer. Prove that 1+2+3.....+(n-1) =0mod n if and only if n is odd.
I assumed first that 1 +2+3...+(n-1)=o mod n. Show n is odd. We know by the Appendix that 1+2+3...+n= [n(n+1)]\2
I plugged n-1 into n in the above equation. which is [n(n-1)]/2. Now do I let that be congruent to 0modn?!?
If you could help I would really appreciate it! Thanks!