I am working on another project relating to Pythagorean triples and came across an unusual property
The primitive triple generator $\left(n, \frac{n^2-1}{2}, \frac{n^2+1}{2}\right)$ for $n\in\mathbb N$ is clearly well known
However, more specifically I want to look at the subset of these such that $n\equiv 5\;(\text{mod }10)$
They can be expressed in the following form:
$$\left(10n+5, 10\sum_{k=0}^nk+2, 10\sum_{k=0}^nk+3\right)$$
I was wondering if this was well known or bore any significance or interesting explanation beyond the trivial
Your formula follows the Pythagorean theorem but, because Pythagorean triples have integer sides, it generates them as non-trivial triples only for odd $n>1$, which happens to include the $ n\in\{5,15,25, ...\}$ that interests you. There is a formula.
\begin{align*} A&=&(2n-1)^2+ & 2(2n-1)k \\ B&=& & 2(2n-1)k+ 2k^2\\ C&=&(2n-1)^2+ & 2(2n-1)k+ 2k^2\\ \end{align*} which generates a superset of what yours does. $$\begin{array}{c|c|c|c|c|} Set_n & k=1 & k=2 & k=3 & k=4 \\ \hline Set_1 & 3,4,5 & 5,12,13& 7,24,25& 9,40,41\\ \hline Set_2 & 15,8,17 & 21,20,29 &27,36,45 &33,56,65\\ \hline Set_3 & 35,12,37 & 45,28,53 &55,48,73 &65,72,97 \\ \hline Set_{4} &63,16,65 &77,36,85 &91,60,109 &105,88,137\\ \hline \end{array}$$ Your formula generates the only the first row where $n=1$ and $k>1.$
If $n=1$, the formula becomes $$A=1+2k\quad B=2k+2k^2\quad C=1+2k+2k^2$$
so this formula will generate your subset plus $(3,4,5)\quad$ if $\quad k\in\mathbb{N}.$
If you let $k\in\{2,7,12\}$ you will generate $$(5,12,13)\quad (15,112,113)\quad (25,312,313)\quad\cdots$$