I'm working on a homework problem that is as follows:
Suppose that $n$ is a positive even integer with $n/2$ odd. Prove that there do not exist positive integers $x$ and $y$ with $x^2 - y^2 = n$.
It looked like a good candidate for proof by contradiction. So I know that I still assume the argument "$n$ is a positive even integer with $n/2$ odd" but will try to show that there exists positive integers $x$ and $y$ with $x^2 - y^2 = n$, and if this reaches a contradiction then I have proven the original conjecture.
So I started with the $n/2$ is odd and rewrote it as $n/2 = 2k+1$, for some $k$ in the integers.
Then I knew that $n = 4k+2$, and then tried so equate that with $x^2 - y^2 = 4k+2$.
EDIT: I then recognized that $x^2 - y^2$ is equivalent to $(x + y)(x - y)$ but that doesn't seem to be very helpful, because if you divide one or the other out you get a term on the RHS in terms of k and x and y.
I'm going to keep playing with it but I don't really have any good strategies going forward. Any help is appreciated!
A more or less mechanical approach is to work modulo $4$. Note that for any integer $k$, $k^2\equiv 0 \pmod{4}$ or $k^2\equiv 1\pmod{4}$.
So, modulo $4$, $x^2-y^2$ can only take on the values $0$, $1$, and $-1$.
More basic, and more useful, is to suppose that $x^2-y^2=n$. Then $(x-y)(x+y)=n$. Note that for any integers $x$ and $y$, the numbers $x-y$ and $x+y$ are both even or both odd. (If we want a proof, their difference $2y$ is even.)
In neither case is $(x-y)(x+y)$ twice an odd integer. You started along these lines. Note that you were one step from the end.
Remark: The reason the second idea is more useful is that when it comes to solving $x^2-y^2=n$, we express $n$ as a product $st$ of integers of the same parity, and solve the system $x-y=s$, $x+y=t$. The solution is $x=\frac{s+t}{2}$, $y=\frac{s-t}{2}$. If $n$ is twice an odd integer, then this process breaks down, because one of $s$ and $t$ will be odd and the other even, so we do not get integers $x$ and $y$.