Is it possible to show (4n^2 + 4n + 1)^m is odd?
I got to this stage in working out a question (i am not sure this stage is right in the first place)
Take a list of odd numbers multiplied: . If this were even then that product would divide by 2 with no remainder. Since no individual odd can be divided by 2, no O_i in the list will divide by it and no O_i can be a multiple of 2, by definition of odd.