Unusual primitive Pythagorean triple identity 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
 A: 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
$$(5,12,13)\quad (15,112,113)\quad  (25,312,313)\quad\cdots$$
A: There's one other formula, for generating Pythagoras triplet.
taking
Hypotenuse = c
side 1 = a
side 2 = b
The formula is
c = m^2 + n^2
b = m^2 - n^2
a = 2mn
where m,n are natural numbers.
this can generate every triplet.
Proof of this is also quite simple, but I am new here so it's hard for me to write that, as I am not getting all the symbols on my laptop keyboard.
