# Thread: Show (4n^2 + 4n + 1)^m is odd

1. ## Show (4n^2 + 4n + 1)^m is odd

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)

2. Originally Posted by sirellwood
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)
Factor 4n^2+4n+1. Then use exponent rules to simplify. The result will show that it is always odd.

3. sorry i dont know what u mean? could you talk me thought it? thanks :-)

4. Originally Posted by sirellwood
sorry i dont know what u mean? could you talk me thought it? thanks :-)
$(2n+1)^2=4n^2+4n+1$ 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.

5. Thanks. that makes perfect sense. Is it ok in a proof to just state that "odd numbers times odd numbers are always odd" or should i expand this a bit further to explain it?

6. Originally Posted by sirellwood
Thanks. that makes perfect sense. Is it ok in a proof to just state that "odd numbers times odd numbers are always odd" or should i expand this a bit further to explain it?
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: $O_1*O_2...O_n$. 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.