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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?

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  • $\begingroup$ There are not too many choices for $c$. It need to be at least $\left\lceil 2000\sqrt{2}\right\rceil = 2829$, must have the form $4k+1$ and when you factor it into prime factors, it cannot contains primes of the form $4k+3$. If I didn't make any mistake, you are leave with $18$ numbers to check: $$2833, 2837, 2845, 2857, 2861, 2873, 2885, 2897, 2909, \\ 2917, 2929, 2941, 2953, 2957, 2965, 2969, 2977, 2993$$ Among them, only $2941^2 = 2059^2 + 2100^2$ fit the bill (by brute-force and by this calculator) $\endgroup$ – achille hui Nov 23 '16 at 20:58
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There are two triples that come close and one that matches your requirements: $$f(14,24)=(2025,2448,3177), GCD(A,B,C)=9$$ $$f(15,21)=(2059,2100,2941), GCD(A,B,C)=1$$ $$f(16,20)=(2201,2040,3001), GCD(A,B,C)=1$$ generated by the functions: $$A=(2n-1)^2+2(2n-1)k$$ $$B=2(2n-1)k+2k^2$$ $$C=(2n-1)^2+2(2n-1)k+2k^2$$

I do not believe others exist, at least not as primitives.

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

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