Proof: m is even or n is odd if and only if (n+1)m^2 is even

[Guest]

Here is my work: I set up 4 cases.
1) if m is even, then (n+1)m^2 is even AND
2) if n is odd, then (n+1)m^2 is even.
3) if (n+1)m^2 is even, then m is even AND
4) if (n+1)m^2 is even, then n is odd.

1st case: Assume m is even, Show (n+1)m^2 is even
I had no trouble with this part of the proof.
2nd case: Assume n is odd, Show (n+1)m^2 is even
I also had no trouble with this.
3rd case: This is where I had trouble
I decided to do a contrapositive proof so
Assume n is even, Show (n+1)m^2 is odd ALSO
Assume m is odd, Show (n+1)m^2 is odd

I started working through this part of the proof and found myself frustrated. Since n is even, by def. n=2k. Then by substitution, (n+1)m^2=(2k+1)m^2. And this is where I stopped. How could I get this in such a form where I can set j= something that will leave (n+1)m^2=2j+1?

Thanks!

 

[pka]

You are making it much too hard.
Suppose \(\displaystyle \left( {n + 1} \right)m^2\) is even then if \(\displaystyle m\) is not even then \(\displaystyle \left( {n + 1} \right)\) must be even so \(\displaystyle n\) is odd. That is it. We are done!