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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} $

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2 Answers 2

up vote 4 down vote accepted

Case(i): If $x, y$ are both even then $z$ is even so $xyz$ is even.

Case(ii): If $x, y$ are one odd on even then $z$ is odd then also $xyz$ is even

Case(iii): If $x, y$ are both odd then $z$ is even so $xyz$ is even

Note: It should be mentioned in the question that $x,y,z$ are integers.

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If either $x$ or $y$ are even, then the product will be even. If not, then $x$ and $y$ are both odd, so $z$ is even.

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