Find all primitive Pythagorean triples such that all three sides are on an interval $[2000,3000]$

Question

For primitive Pythagorean triples $(a,b,c)$, the following is valid: $$a=m^2-n^2,b=2mn,c=m^2+n^2$$ or $$b=m^2-n^2,a=2mn,c=m^2+n^2$$

$gcd(m,n)=1,m>n$

If numbers $a,b,c$ are relatively prime, then $(a,b,c)$ is a primitive Pythagorean triples, otherwise it is not primitive.

For all Pythagorean triples (not only for primitive, where $gcd(a,b,c)=1$), the following is valid: $$(d(m^2-n^2))^2+(2dmn)^2=(d(m^2+n^2))^2$$

If we would have one fixed side, then we would factorize the it's size and consider the number of cases which is equal to the number of factors in the fixed side.

Here, we are given that every side is on an interval $[2000,3000]$.

What is the method to find all primitive Pythagorean triples in this case?

Answer 1

The only primitive triple that meets your requirements is: $$f(n,k)=f(15,21)=(2059,2100,2941), GCD(A,B,C)=1$$ generated using equations I developed in a spreadsheet:

$$A=(2n-1)^2+2(2n-1)k$$ $$B=2(2n-1)k+2k^2$$ $$C=(2n-1)^2+2(2n-1)k+2k^2$$

I do not believe others exist as primitives. To exist, the ratio of the hypotenuse to the smallest leg must be $1.5:1$ or less. Then some integer multiple of the shortest leg must be greater than $2000$ and the same integer multiple of the hypotenuse must be less than $3000$.

for non-primitives you have $$f(2,2)=(21,20,29)\text{ times 100}$$ $$f(4,5)=(119,120,169)\text{ times }17$$ $$f(6,7)=(275,252,373)\text{ times } 8$$ $$f(6,8)=(297,304,425)\text{ times } 7$$ $$f(8,10)=(525,500,725)\text{ times }4$$ $$f(9,12)=(697,696,985)\text{ times }3$$ $$f(10,15)=(931,1020,1381)\text{ times }2$$