$$(2n + 1)^2 + (2n^2 + 2n)^2 = (2n^2 +2n +1)^2$$
It can be used to generate infinitely many sides of right-angled triangles with integer lengths by putting values of $n = 1, 2, 3, ... $
I wanted to know that how we came to this equation. How do we know that putting n = 1, 2, 3, … into this we'll get all the sides of a right-angled triangle. I'm trying to find more about this on the internet, if you can help me what should I find about.
On
Your formula notes that often Pythagorean Triples are of the form $(t_1, t_2, t_2+1)$. It attempts to see when this is the case and this formula generates that, we have: $$(2n^2+2n+1)^2-(2n^2+2n)^2=4n^2+4n+1=(2n+1)^2$$
Since the difference here is just a square, it's a valid triple. However, this misses many triples, particularly multiples of existing triples, but an example is $(8,15,17)$. I prefer the usage of:
$$(p^2-q^2)^2+(2pq)^2=(p^2+q^2)^2$$ This is most notable for its link to complex numbers, it is $$\Re(z^2)+\Im(z^2)=|z^2|$$ and this can be used to generate them. Set your calculator to Complex mode and use $(1000\text{Ran#}+\text{1000Ran#}i)^2=$ and every result will be the base and height of a Pythagoras triangle.
(Ran# is my calculators random number generator, it might be different to yours. Also it generates from $0.001$ to $1$, max $3$dp, hence my multiplying by $1000$)
On
Consider the Primitive Pythagorean triple $(a, b, c)$. Consider $b=4T_n$ where $T_n$ is the $n^{\text{th}}$ triangular number. Notice that if this is so, then the value of $c$ is $4T_n+1$.
Note that the formula for the $n^{\text{th}}$ triangular number is given by $\frac{n(n+1)}{2}$. All that we are left to do is solve for $a$ from the following equation: $a^2+[4T_n]^2=[4T_n+1]^2$. Solving this gives the value of $a$ as $2n+1$. Hence, $(a, b, c)=(2n+1, 4T_n, 4T_n+1)$ is a Pythagorean triple.
On
Your formula (I call it "yours" in reference to the post) is a special case of a formula I developed in 2009 and have occasionally posted here in answer to questions about Pythagorean triples.
\begin{equation}\qquad A=(2n-1)^2+2(2n-1)k\qquad B=2(2n-1)k+2k^2\qquad C=(2n-1)^2+2(2n-1)k+2k^2 \end{equation}
It generates the subset of triples where $GCD(A,B,C)$ is an odd square. The table below shows these sets where $n$ is the set number and $k$ is the "count" or member number within the set.
\begin{array}{c|c|c|c|c|c|c|} n & k=1 & k=2 & k=3 & k=4 & k=5 & k=6 \\ \hline Set_1 & 3,4,5 & 5,12,13& 7,24,25& 9,40,41& 11,60,61 & 13,84,85 \\ \hline Set_2 & 15,8,17 & 21,20,29 &27,36,45 &33,56,65 & 39,80,89 & 45,108,117 \\ \hline Set_3 & 35,12,37 & 45,28,53 &55,48,73 &65,72,97 & 75,100,125 & 85,132,157 \\ \hline Set_{4} &63,16,65 &77,36,85 &91,60,109 &105,88,137 &119,120,169 & 133,156,205 \\ \hline Set_{5} &99,20,101 &117,44,125 &135,72,153 &153,104,185 &171,140,221 & 189,180,261 \\ \hline \end{array} If we let $n=1$, my formula becomes $\quad A=(2k + 1)^2\quad B=(2k^2 + 2k)^2\quad C=(2k^2 +2k +1)^2\quad$ which matches yours exactly and generates the sub-subset $(Set_1)$ in the table above.
How developed? Notice that the difference $(d)=(C-B)=(2n-1)^2$ and the increment $(i)$ between values of $A$ is $2(2n-1)$. Now notice that $A$ is the sum of $(d)$ plus the increment $(i)$ times a multiplier $(k)$. From here, the B-function and C-function follow by substituting the now-known expressions for $A$ and $(C-B)$ into the Pythagorean theorem.
If you start with the sequence of square numbers, and take all the differences between consecutive terms, you get all the odd numbers in sequence. For instance, $$ 2^2 - 1^2 = 3\\ 3^2 - 2^2 = 5\\ 4^2-3^3 = 7 $$ Some times, that odd number happens to be a square itself. For instance, we have $$ 5^2 - 4^2 = 9 = 3^2\\ 13^2 - 12^2 = 25 = 5^2 $$ Rearranging these, we get Pythagorean triples: $$ 3^2 + 4^2 = 5^2\\ 5^2 + 12^2 = 13^2 $$ If we want to describe all the different Pythagorean triples that appear this way, we end up with exactly your expression.