I needed to prove the following statement: $$\forall\{x, y, z\}\subset\mathbb{Z}, \ \bigg(x + \frac{z}{2(x-y)}\bigg)^2 - \bigg(y + \frac{z}{2(x-y)}\bigg)^2\in \mathbb{Z} \tag1$$
From the well-known formula, $$a^n - b^n = (a-b)\sum_{k=1}^n a^{n-k}b^{k-1}$$ I know that $x - y\mid c$ such that $c$ is the difference of the two squares. (I substituted it as $c$ just for simplicity.) I also know that since $x$ and $y$ are integers, we have that $x - y$ is an integer, therefore $c$ must also be an integer? Is this proof valid? It just looks too simple, but I believe it is because I knew of the formula anyway, even though the formula is not too difficult to prove.
Assume that $a^n - b^n = d(a - b)$ for some $d \in \mathbb{Z}$, then we have as follows: $$\begin{align} d &= \frac{a^n - b^n}{a - b} \\ \\ &= \frac{a^{n - 1}(a - b) + ba^{n - 1} - b^n}{a - b} \\ \\ &= a^{n - 1} + b\bigg(\frac{a^{n -1} + b^{n - 1}}{a - b}\bigg) \end{align}$$ And now with some simple repetition, we obtain our desired formula.
But, if the proof of $(1)$ is correct, then just out of curiosity, is there another way to prove it? I guess that's two questions, then.
Thank you in advance.