Can we simplify a proof of parity equality?

I want to prove that the following statements are equivalent: “m−nm-n is even, m2−n2m^2-n^2 is even, and m2+n2m^2+n^2 is even. I know that given a number xx, xx & x2x^2 have the same parity. So is it sufficient to just prove that the sum is even if they have the same parity?

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Note that for all natural numbers m,nm,n, it holds that m2+n2=(m2−n2)+2n2=(m−n)(m+n)+2n2.m^2+n^2=\left(m^2-n^2\right)+2n^2=(m-n)(m+n)+2n^2.
– Git Gud
Oct 20 at 20:53

  

 

@GitGud I’m not sure I follow what you are pointing out… this is still even if m & n share the same parity
– Bryyo 357
Oct 20 at 20:58

  

 

The equalities tell you that m2+n2,m2−n2m^2+n^2, m^2-n^2 and (m−n)(m+n)(m-n)(m+n) have the same evenness. To get that m−nm-n has the same evenness as the above, consider how the evenness of m−nm-n relates to the evenness of mm and nn.
– Git Gud
Oct 20 at 21:05

  

 

@GitGud m and n would have to have the same evenness, too
– Bryyo 357
Oct 20 at 22:34

  

 

That’s correct. Can you formalize the argument?
– Git Gud
2 days ago

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