How to show that xyz is even( 2|xyz), when $x+y=z$? for example if x=5, y=12 then z=17. And $2\mid 5 \cdot 12 \cdot 17 $ <=> $2\mid 5 \cdot 2 \cdot 6 \cdot 17 $ ok.
One way to show that zyz is even is to make an array and use the knowledge that case $x \cdot y$ is odd and $z$ is odd never happens. Latter comes from the case that if x odd and y odd then $x\cdot y$ is odd. And if x even y odd or x even y odd then z odd. So here is contradiction, so z can never be odd. Am I right? $ \begin{array}{ c| c } * & z \quad \text{even} & z \quad \text{odd} \\\hline x \cdot y \quad \text{even} & even & even \\ x \cdot y \quad \text{odd} & even & odd \end{array} $