Reciprocals by 2

$We\ have,$ $\frac{1}{n-1}+\frac{1}{n+1}=\frac{p}{q}$ $By\ solving,$ $\frac{2n}{n^2-1}=\frac{p}{q}$ $p^2+q^2=\left(2n\right)^2+\left(n^2-1\right)^2$ $p^2+q^2=4n^2+n^4+1-2n^2$ $p^2+q^2=n^4+1+2n^2$ $p^2+q^2=\left(n^2+1\right)^2$ $By\ square\ rooting\ both\ the\ sides,$ $\sqrt{p^2+q^2}=\left(n^2+1\right)\$ $Hence\ p^2+q^2\ is\ perfect\ square\ always$

@Chung Kevin : Here's your pythagorean triplet approach.

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$\begin{aligned} & \dfrac{1}{n-1} + \dfrac{1}{n+1} &=& \dfrac{p}{q} \\ \implies & \dfrac{2n}{n^2-1} &=& \dfrac{p}{q} \end{aligned}$

A pythagorean triplet is of form:

$(2mn, n^2-m^2, n^2+m^2)$

Putting $m=1$ we get,

$(2n, n^2-1, n^2+1)$

Since the term $n^2+1$ is the largest and so taking it as hypotenuse of a right triangle, with angle $\Phi$ opposite to side $2n$ we get,

$\tan \Phi = \dfrac{2n}{n^2-1}$

A similar right triangle can be constructed with angle $\Phi$ opposite to the side $p$ .

Then,

$\tan \Phi = \dfrac{p}{q}$

So, $p$ and $q$ are also sides of a right triangle containing the right angle.

Thereby, we conclude, by pythagoras theorem, that there exists some $k$ which is an integer such that $k^2 = p^2 + q^2$ , hence coming to the conclusion that $p^2+q^2$ is always a perfect square.

Note that this works for $|n| \ge 2$ , where $n \in \mathbb{Z}$ since there is no defined triplet for $0 \le |n| \le 1$ , noting that $n \in \mathbb{Z}$ .

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I think you should have mentioned $|n| \ge 2$ . Although this is not a problem since $p$ and $q$ are positive. But just for clarity.