Why $(2m)^2 + (m^2 - 1)^2 = (m^2 + 1)^2$ results in pythagorean triples? As you increase the value of n, you will generate all pythagorean triples whose first square is even. Is there any visual proof of the following explicit formula and where does it come from or how to derive it?
$(2n)^2 + (n^2 - 1)^2 = (n^2 + 1)^2$




$(2n)^2+(n^2-1)^2=(n^2+1)^2$
$(2n)^2+(n^2-1)^2=(n^2+1)^2$
$(2n)^2+(n^2-1)^2=(n^2+1)^2$




$(2*0)^2+(0^2-1)^2=(0^2+1)^2$
$(2*1)^2+(1^2-1)^2=(1^2+1)^2$
$(2*2)^2+(2^2-1)^2=(2^2+1)^2$


$(2*0)^2+(0-1)^2=(0+1)^2$
$(2*1)^2+(1-1)^2=(1+1)^2$
$(2*2)^2+(4-1)^2=(4+1)^2$


$0^2+1^2=1^2$
$2^2+0^2=2^2$
$4^2+3^2=5^2$


$0+1=1$
$4+0=4$
$16+9=25$


$1=1$
$4=4$
$25=25$




All Pythagorean triple have a "leg" that is a multiple of
$\space4.\space$ Your formula generates a subset of these.
We begin with Euclid's formula shown here as
$$A=m^2-k^2\quad B=2mk \quad C=m^2+k^2$$
If we let $\space k=1,\space$ we have
$\quad A=m^2-1\quad B=2m\quad C=m^2+1$
This formula also generates trivial triples where one term is zero. A variation of this replaces $\space m\space$ with $\space(2n-1+k)=2n\space$ and produces a formula that generates non-trivial triples for any natural number $n$.
$$\quad A=4n^2-1\quad B=4n\quad C=4n^2+1$$
By inspection, we can see that the middle term will always be even and that $\space C-A=2\space$ for all values of
$\space n.\space$ Algebraically below, we can also "see" that the formula is valid.
\begin{align*}A^2+B^2=&(4n^2-1)^2+(4n)^2\\
=&(16 n^4 - 8 n^2 + 1)+(16n^2)\\
=&(16 n^4 + 8 n^2 + 1)\\
=&(4n^2+1)^2=C^2\\
\end{align*}
