Dickson's method for generating pythagorean triples can not find all triples - Known or not? I recently came across Dickson's method on Wikipedia for generating pythagorean triples. I implemented this in a computer programming language and found some oddities, which was based on the fact that there was missing triples, ie, the Dickson method did not generate all of the triples.

The specific example that I found was the triple $(65, 72, 97)$ which satisfies $65^2(=4225) + 72^2(=5184) = 97^2 (=9409)$ ie it is a valid pythagorean triple, but it can not be found with the Dickson method:

In the Dickson method $a = r + s$, $b = r + t$ and $c = r + s + t$ where $r$ and $s$ are factors of $r^2/2$. In this specific case, $r=40$ which implies $s = 25$ and $t = 32$ with neither of $s$ and $t$ being factors of $40$.

My question is whether this is a known fact or not? If it is a known fact, is there a source somewhere that I can cite to fix the Wikipedia page which claims it finds all triples?
Here is a proof that Dickson's method works. Let $a, b, c$ be three integers. Set
$$r = a + b - c$$
$$s = c - b$$
$$t = c - a$$

which are manifestly integers if and only if $a, b, c$ are integers. Now we compute that
$$r^2 - 2st = a^2 + 2ab + b^2 - 2ac - 2bc + c^2 - 2c^2 + 2ac + 2bc - 2ab = a^2 + b^2 - c^2$$

hence $r^2 = 2st$ if and only if $a^2 + b^2 = c^2$.
Here is a better solution which I have formulated.
$$a = s^2 + 2sn                ..................       \qquad\rm Formula\ 1.J$$
$$b = 2n^2 +2sn              ....................       \qquad\rm Formula\ 1.K$$
$$c = s^2 +2n^2 +2sn         ....................        \qquad\rm Formula\ 1.L$$

This will generate all triples.