Is the prove correct for: If both ab and a + b are even then both a and b are even Show:
If both $ab$ and $a + b$ are even, then both $a$ and $b$ are even
Proof:
Assume both $ab$ and $a + b$ are even but both $a$ and $b$ are not even
Case1: one is odd
$a=2m+1$, $b=2n$
Hence $a+b = (2m+1) + 2n = 2(m+n) + 1$
Case2: both are odd
$a=2m+1$, $b=2n+1$
Hence $ab = (2m+1)(2n+1) = 2(2mn+m+n) + 1$
Therefore both $a$ and $b$ have to be even for both $ab$ and $a+b$ to be even.
My question is that 
1) Is this proof correct? 2) Is this proof by contradiction or negation or any other?

You might also argue that as $a + b$ is even, both are odd or both are even. But if $a$ and $b$ are odd, then $a b$ is odd, contradicting the premises. So both are even.
