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