Another proof for an infinite number of Pythagorean triples I’m not sure if this has been mentioned before (and I truly apologize if someone thought about it already) , but I tried to adopt a geometrical approach for the proof for an infinite number of Pythagorean triplets. Please read it once and please point out any mistakes . 
Any Pythagorean triplet can be expressed in the form of $x^2+ y^2=r^2$ (r being an integer). Keeping $r$ constant , we can easily conclude that the equation is that of a circle. Rearranging the equation gives $y= \sqrt{r^2-x^2}$. (considering only natural numbers) . Now the domain of $x$ is $-r<x<r$ . Considering the base case , $r=1$ there is at least one integer pair that satisfies its equation , it being $(0,1)$. For $r>1$ , any integral value of $x$ gives a corresponding integral value of $y$, while the condition $-r<x<r$ still holds true. As there are infinite number of natural numbers $r$ , the result follows. 
I would really like to know if I’ve missed on something so please guide me . Thank you ! 
 A: Your statement is correct but your proof is not.
For example, consider the equation $$y= \sqrt {16-x^2}$$
You have claimed that for any integral value of $x$ you get an integral value of $y$
For $x= 1, 2, 3$ you get $y= \sqrt {15} , \sqrt {12}, \sqrt {7} $ and none of these numbers are integers. 
The  only integers that we get out of that are $0$ and $4$ which are trivial solutions to $ x^2+y^2=16$
A: The following functions are original and may serve your purpose. They look nothing like Euclid's formulas because they were developed from an observation [of 8 million spreadsheet formulas] that all triplets where $GCD(A,B,C)=(2n-1)^2,A,B,C,n\in \mathbb{N}$. This includes all primitives. The main value in this proof is that it shows there is a unique triple for ever pair of natural numbers, $n$ and $k$.
Theorem:
$$\forall n,k \in \mathbb{N}, \exists A,B,C\in \mathbb{N}:A^2+B^2=C^2\text{ if } A=(2n-1)^2+2(2n-1)k$$
Proof: Let $$A=(2n-1)^2+2(2n-1)k$$
Solving $A^2+B^2=C^2$ for $B$ and $C$, respectively and substituting $A$,  we find that
$$B=2(2n-1)k+2 k^2$$and$$C=(2n-1)^2+2(2n-1)k+2k^2$$
We can then show that
$$A^2=(2n-1)^4+4(2n-1)^3 k+4(2n-1)^2 k^2$$
$$B^2=4(2n-1)^2 k^2+8(2n-1) k^3+4k^4$$
$$C^2=(2n-1)^4+4(2n-1)^3 k+8(2n-1)^2 k^2+8(2n-1) k^3+4k^4$$
$$A^2+B^2=(2n-1)^4+4(2n-1)^3 k+8(2n-1)^2 k^2+8(2n-1) k^3+4k^4=C^2$$
$\therefore \forall n,k \in \mathbb{N},\exists A,B,C\in \mathbb{N}:A^2+B^2=C^2 \text{ if } A=(2n-1)^2+2(2n-1)k\text{ } \blacksquare$
