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


2 Answers 2


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

  • 1
    $\begingroup$ 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 $\endgroup$ 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.

  • $\begingroup$ 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\$ $\endgroup$
    – poetasis
    Jul 20, 2021 at 0:43

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