To get all the (b,a,a+1) Pythagorean triplets

Question

In $(b,a,a+1)$ I found that $a+a+1=b^2$, in which $a>b$

You guys probably knew this but I randomly thought this thing in $5,12,13$ and then found it in all with $b,a,a+1$

So anyways That means $2a+1=b^2$ If we take b as any odd number and the reason for odd is so that we can get $b^2$ as odd number and then $b^2-1$ be a even number and as $b^2-1=2a$, we will get a integer $a$

So, $b = 2n+1$ so for each value of $n$, we get a triplet and yeah it's cool to have all triplets of this form

Eg.

1. $n = 5$: $b = 11$
   $2a+1 = 121$
   $a = 60$
   So the triplet is $(11,60,61)$
2. $n = 8$: $b = 17$
   $2a+1 = 289$
   $a = 144$
   So the triplet is $(17,144,145)$

My question is I can't be anywhere near smart just by studying high school maths, and this isn't a hard find. So why this all was never taught to us yet.And sorry if this was very dumb...

Answer 1

I developed a formula in 2009 that generates sets of Pythagorean triples where $\,C-B=(2n-1)^2\,$ for each set. If you let $\,n=1,\,$ then you get the triples in $\,Set_1\,$ where $\,C-B=1^2.\quad$ Side-$A\,$ is always of the form $\,2x+1\,$ so it can be any odd number greater than $\,1\,$ and $\,Set_1\,$ includes all of these.$\quad$ Side-$B\,$ is always of the form $\,4x.\quad$ Side-$C\,$ is always of the form $\,4x+1.$ \begin{align*} &A=(2n-1)^2+2(2n-1)k\\ &B=\phantom{(2n-1)^2+{}} 2(2n-1)k+2k^2\\ &C=(2n-1)^2+2(2n-1)k+2k^2 \end{align*}

$$\begin{array}{c|c|c|c|c|c|c|} n & k=1 & k=2 &k=3 & k=4 & k=5\\ \hline Set_1&3,4,5 &5,12,13&7,24,25&9,40,41&11,60,61\\ \hline Set_2&15,8,17&21,20,29 &27,36,45 &33,56,65&39,80,89\\ \hline Set_3&35,12,37&45,28,53&55,48,73&65,72,97&75,100,125\\ \hline Set_{4}&63,16,65&77,36,85&91,60,109&105,88,137&119,120,169\\ \hline Set_{5}&99,20,101&117,44,125&135,72,153&153,104,185&171,140,221\\ \hline \end{array}$$

On the other hand, if you want all triples where $\,B=A\pm 1,\,$ you should use Euclid's formula $\quad A=m^2-k^2\quad B=2mk\quad C=m^2+k^2\quad $

and derive the parameters from Pell numbers where

$$ m_n= \frac{(1 + \sqrt{2})^{n+1} - (1 - \sqrt{2})^{n+1}}{2\sqrt{2}}\qquad k_n= \frac{(1 + \sqrt{2})^n - (1 - \sqrt{2})^n}{2\sqrt{2}}$$

For example

\begin{align*} m_1&= \frac{(1 + \sqrt{2})^{2} - (1 - \sqrt{2})^{2}}{2\sqrt{2}}=2 & k_1= \frac{(1 + \sqrt{2})^1 - (1 - \sqrt{2})^1}{2\sqrt{2}}=1\\ & F(2,1)=(3,4,5)\\ \\ m_2&= \frac{(1 + \sqrt{2})^{3} - (1 - \sqrt{2})^{3}}{2\sqrt{2}}=5 &k_2= \frac{(1 + \sqrt{2})^2 - (1 - \sqrt{2})^2}{2\sqrt{2}}=2\\ & F(5,2)=(21,20,29)\\ \\ m_3&= \frac{(1 + \sqrt{2})^{4} - (1 - \sqrt{2})^{4}}{2\sqrt{2}}=12 &k_3= \frac{(1 + \sqrt{2})^3 - (1 - \sqrt{2})^3}{2\sqrt{2}}=5\\ & F(12,5)=(119,120,169)\\ \\ m_4&= \frac{(1 + \sqrt{2})^{5} - (1 - \sqrt{2})^{5}}{2\sqrt{2}}=29 &k_4= \frac{(1 + \sqrt{2})^4 - (1 - \sqrt{2})^4}{2\sqrt{2}}=12\\ & F(29,12)=(697,696,985) \end{align*}

Answer 2

This is a very good find. It is known, but being able to discover things on your own is an achievement in itself.

Here's something more for you--spot the pattern:

$$3^2+4^2=5^2$$ $$5^2+12^2=13^2$$ $$7^2+24^2=25^2$$ $$9^2+40^2=41^2$$ $$11^2+60^2=61^2$$ $$13^2+84^2=85^2$$

$$\vdots$$