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)
$\displaystyle (2n+1)^2=4n^2+4n+1$ A power raised to another power results in multiplying the two so the final simplification is (2n+1)^(2m). For any n, 2n+1 is odd. Odd numbers times odd numbers are always odd, no matter how many, so for any m the result is still odd.
I think it's simple enough to not need proof but if you want to it's not hard.
Take a list of odd numbers multiplied: $\displaystyle O_1*O_2...O_n$. 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.