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And similarly the sum and difference of two odd numbers is even, but the product of odd numbers is odd

Can someone help me out here? I'm really stuck on how to start this proof after making my theorem. Thank you!!

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    What are your thoughts on the problem? Have you tried to approach it on your own? – 2017-02-13
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    An odd question ? – 2017-02-13
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    I tried approaching it by using 2m+1 since that is an odd number and i would just start doing 2(2m+1) and Im not even sure. cause we have to find the difference and i really wasnt sure about it – 2017-02-13

2 Answers 2

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if you have two even numbers, say $$2m,2n$$ with integers $m,n$ then the sum is given by $$2n+2m=2(m+n)=2k$$ thus our sum is even, where we have $$k=m+n$$ and $k$ is an integer number. analogously we have $$2m-2n=2(m-n)=2k$$ where $k=m-n$ is an integer number. and last but not least we have $$2n\cdot 2m=2(2mn)=2k$$ where $k=2mn$

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Hint: If $a$ and $b$ are even numbers, then they are of the form $a = 2n$ and $b = 2m$ for $m,n$ integers.