<s>[INST] 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?
$(2n)^2 + (n^2 - 1)^2 = (n^2 + 1)^2$




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




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


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


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


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


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



 [/INST] The formula that you are referring is a sub-case of the  Euclid's formula
Accroding to the  Euclid's formula it is true that:
$$\text{Given an arbitrary pair of integers m and n with m$\gt$n and m,n$\gt$0}$$
$$a=m^2-n^2 \text{ , } b=2mn \text{ , } c=m^2+n^2 \text{ form a pythagorean triple}$$
$\text{In your case for:} n=1 \text{ and } m\epsilon\mathbb{N} $

You can found more about this  forumla in this links:
proof of euclid's formula MathExchange 
Pythagorean triple-wikipedia
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