Order in direct proofs with even numbers I'm doing an advanced maths class for high school and we have just started a topic about proofs.
One of the questions (assume all numbers are integers here) is to prove that if $x\cdot y$ and $x + y$ are even, then both $x$ and $y$ are even. 
I know that an even number is defined by 2m. With a proof like this, could I start my proof by assuming that x and y are even and then substituting into $x\cdot y$ and $x + y$? Or should I be starting from them and proving that $x$ and $y$ are even.
Alternatively, would it be better (or simpler) to prove starting by assuming that the numbers are all odd? 
 A: There are at least two simple ways to prove what you want to prove.
The most basicway to prove what you want to prove is to separate four possible cases:


*

*$x$ and $y$ are both odd

*$x$ is odd and $y$ is even

*$x$ is even and $y$ is odd

*$x$ and $y$ are both odd.


If you can prove that cases $1$, $2$ and $3$ are impossible, then you have proven that $4$ is true.

Another way to prove is to first use the fact that $x+y$ is even. You can probably easily prove that if $x+y$ is even, then either $x$ and $y$ are both odd or they are both even. This is simple to prove since its equivalent statement, the statement "if one of $x,y$ is odd and the other is even, then $x+y$ is odd" is simple to prove.
Once you have shown that $x$ and $y$ are both odd or both even, you just have to show that they cannot both be odd, which is simple since if they are both odd, then $xy$ is odd.

You can also start from the other end and use the fact that $xy$ is even, since that means that at least one of the numbers is even. Then you prove that the other cannot be odd as that yould mean that $x+y$ is odd.
