# Identifying Proof Method and Implementing It

The question I am working on is:

Prove that if $m+n$ and $n+p$ are even integers, where $m$, $n$,and $p$ are integers, then $m+p$ is even. What kind of proof did you use?

I was thinking--and I aware that this may not be the most efficient method--of proving four different cases: $m$ and $p$ are both even; $m$ and $p$ are both odd; or $m$ and $p$ are opposite parity. Would this work? Or is there a better way?

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It should work. –  A.P. Feb 12 '13 at 18:05
What you mention should work perfectly. As for the name of the method, I don't know, but I've seen that sometimes be called prove by inspection. Which is what you do when you analise all possible cases... –  MyUserIsThis Feb 12 '13 at 18:32

Let $m+n=2a,n+p=2b$ where $m,n,a,b$ are integers

$\implies m+n+n+p=2(a+b)\implies m+p=2(a+b-n)$

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Oh, that's clever. However, I'd like to know if my method is valid. –  Mack Feb 12 '13 at 18:03
@EliMackenzie, it will definitely work. But, playing with all the combinations of parities is error prone. –  lab bhattacharjee Feb 12 '13 at 18:05

This is quite easy way : $(m+p)=((m+n)+(n+p))-2n$

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Hint $\,$ If $\rm\,m\,$ and $\rm\,p\,$ have same parity as $\rm\,n\,$ then $\rm\,m+p\,$ has same parity as $\rm\,n+n,\,$ which is even.

Said in congruence language: $\rm\ mod\ 2\!:\ m\equiv n,\ p\equiv n\:\Rightarrow\: m+p\equiv n+n\equiv 2n\equiv 0$

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