How to Prove Pythagorean Triple Formula I'm having a hard time finding a proof for how they derived the Pythagorean triple formula. 


*

*It's hard to find the proof online and 

*When I do find it, it's hard to understand.


$a = p^2 − q^2 , b = 2pq, c = p^2 + q^2$
Please explain simply. 
 A: The starting point is observe that: $b^2 = c^2-a^2 = (c-a)(c+a)$. Thus if $b$ is even, then since $(c+a) - (c-a) = 2a$ which is even. So $c-a,c+a$ are both even. Then you can put $c-a = 2p^2, c+a = 2n^2$. If $b$ is odd, then both $c-a, c+a$ are odd. Thus you can write $c-a = k^2, c+a=r^2$. In both cases,  you are led to the solution above. 
A: Deriving Euclid's function for generating Pythagorean triples:
A Visual Pythagorean Demonstration question got answers about how a picture probably prompted the idea that $A^2+B^2=C^2$ but how do you come up with equations for $A,B,\text{ and }C$?
In any Pythagorean triple, if it is not an even multiple of a primitive, one leg is odd, one leg is even and the hypotenuse is odd.
Let's begin with the even side and experiment. For example $(3,4,5)$ we have $2X$ something and the only other $somethings$ it could be are $2$ and $1$. We might notice that $2^2-1^2=3$ and that $2^2+1^2=5$ and that it looks to be worth exploring. Our inputs for this 2-input function were an even number and an odd number (2,1) and we can't be any smaller than $(3,4,5)$ so let's try larger. 
If we let $A=2^2-2^2=0$, we have a trivial beginning, at best, and an even number for $A$ if we use both odd or  even inputs $(m,n)$.  For $f(3,2): $ $A=3^2-2^2=5$, $B=2*3*2=12$, and $C=3^2+2^2=13$?
It seems to work; now we can try to prove it.
$$A=m^2-n^2\Rightarrow A^2=m^4-2m^2n^2+n^4$$
$$B=2mn\Rightarrow B^2=4m^2n^2$$
$$C=m^2+N^2\Rightarrow C^2=m^4+2m^2n^2+n^4$$
$$A^2+B^2=(m^4-2m^2n^2+n^4)+(4m^2n^2)=m^4+2m^2n^2+n^4=C^2$$
Now we have proof that the [Euclidean] function works for all natural numbers, though we may see trivials such as f(1,1)=(0,2,2) and $3^{rd}$ quadrant triples such as $f(1,2)=(-3,-4,5)$. We started with the smallest triple, experimented with a simple pattern, found it worked for another triple and then proved that it holds for all $A^2+B^2=C^2$. That's about all we can do but I think we've done a lot here. 
A: You can start by comparing the natural numbers in the simplest manner: $a < b < c < ... < N$  , but their differences will be trivial $(c-b= b-a=1)$, so you can square them all without losing their order, because they're all positive integers, and their differences will become larger.
Now ask yourself: how can I use these numbers in equations when they are bounded by inequality signs? The answer is to add some number $n$ to the lowest term to bring the next one into an equation like this:
$$n+a^2=b^2<c^2<...<N^2$$
So that you can work with $n=b^2-a^2$ by discarding the others. Don't worry, it'll work for all of them, but now $n$ needs to be defined in terms of $b^2$ and $a^2$. They represent a difference of squares, which is, most of the time, not a square unless those $a$ and $b$ are two of the Pythagoren Triples, and that gives us a clue of what to do next: square both sides to have those terms in the same degree. I'll proceed quicker now:
$$n^2=(b^2-a^2)^2
=(b-a)^2(b+a)^2
=(b^2+a^2-2ab)(b^2+a^2+2ab)
=(b^2+a^2)^2-(2ab)^2$$
Now replace $n^2$ with this result when it's equal to that difference of squares, make the necessary transformations for addition, and you get:
$$(b^2-a^2)^2+(2ab)^2=(b^2+a^2)^2$$
This looks like the Pythagoras Theorem, because the Pythagorean Triples have to satisfy that equation, but:
$$p=b^2-a^2$$
$$q=2ab$$
$$r=b^2+a^2$$
... are only those triples when $a$ and $b$ are not multiples of each other, except for the trivial case when they're all multiples of 1, which means that they can be coprime. Should both be even, the expression has a common factor, and they're not in reduced form. That's a proof for later.
