Prove that if $ (x+y)$ is even, then $(x−y)$ is even, for integers. Did I prove correctly? is this a direct proof?

Question

For $x+y$ to be even, either $x$ and $y$ are both even, or $x$ and $y$ are both odd.

If $x$ and $y$ are both even we obtain: $x=2k$ and $y=2j$. substituting into $x-y$ we get $2k-2j$. $2(k-j)$ is even, so we have proved the first case.

Now if $x$ and $y$ are both odd, we obtain: $x=2k+1$ and $y=2j+1$. substituting into $x-y$ we get $(2k+1)-(2j+1)$, $2k+1-2j-1= 2k-2j= 2(k-j)$ which is even, and we have proved the remaining case.

Answer 1

Actually, you can have a direct proof that $x+y$ and $x-y$ have the same parity, since $(x+y)-(x-y)=2y$ is even.

Answer 2

Your proof (by cases) is correct, but you could do simpler without considering cases: if $x+y = 2q$, then $x-y = (x+y)-2y = 2q - 2y$ so $x-y = 2(q-y)$, which means that $x-y$ is even.

Answer 3

Note that if the difference between $p^2$ and $q^2$ is a multiple of $4,$ then $p$ and $q$ must be of equal parity, for otherwise, we have wlog that $(2m+1)^2-(2n)^2=(2m+1-2n)(2m+1+2n)=(2M+1)(2N+1),$ which is never even.

Now since $(x+y)^2-(x-y)^2=4xy,$ it follows that $x+y$ and $x-y$ are of equal parity, and in particular if one of them is even then the other must be as well.