# 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

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 $$(5,12,13)\quad (15,112,113)\quad (25,312,313)\quad\cdots$$

• Thanks for taking the time to make this! Table definitely helped me with the larger problem I’m working on, gotta do some coding from here Jul 19, 2021 at 9:07

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.

• What you have demonstrated is Euclid’s formula, the gold standard for generating Pythagorean triples. The drawback(s) with it are that it generates trivial triples like $(0,2,2)$ and doubles or square multiples of primitives. My formula generates no trivials, limits non-primitives to odd-square multiples like $(27,36,45)$, and makes it easy to see that the OP formula generates only the members of $Set_1$ from my formula. As for characters on the keyboard, you need only to learn Tex, of which MathJax is a variation. For instance, your equation would look better is you used \$A^2+B^2=C^2\$ Jul 20, 2021 at 0:43