An interesting tweet, some time ago, from @RJS2212:

And of course, you wonder two things: a) why does it work, and b) can all Pythagorean triples be written that way?

The first one is an easier proposition, and I’ll set it up like so: we have $\frac{1}{n-1} + \frac{1}{n+1} = \frac{2n}{n^2-1}$.

Are these the legs of a Pythagorean triangle? Let’s see, is $(2n)^2 + (n^2 - 1)^2$ a square? It expands as $4n^2 + n^4 - 2n^2 + 1 = n^4 + 2n^2 + 1 = (n^2 + 1)^2$, which is clearly a square.

Now, though, are all Pythagorean triples of this form — that is, can they be written as $(2n, n^2 -1, n^2 + 1)$? Clearly not — for example, $(9, 40, 41)$ isn’t in this form; however, it can be written as the sum of two unit fractions with denominators differing by two ($\frac{9}{40} = \frac{1}{8} + \frac{1}{10}$).

How about $(20, 21, 29)$? $\frac{20}{21}$ would sensibly be split up as $\frac{a}{3} + \frac{b}{7}$. If $7a + 3b = 20$, $a = b = 2$ is a solution. Alternatively, $\frac{20}{21} = \frac{1}{\frac{3}{2}} + \frac{1}{\frac{7}{2}}$ — unit fractions (of a sort) where the denominators differ by 2.

Is it the case that any pair of unit fractions with rational denominators differing by 2 gives a Pythagorean triple? (I suspect so). Can all triples be generated this way? (I suspect so, too).

The proofs, though, are left as an exercise.