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?

  • $\begingroup$ 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) $\endgroup$ Nov 23, 2016 at 20:58
  • $\begingroup$ @user300047 I edited my answer to include an explanation of where these functions came from. My posts that include them almost always get downvoted but I would like you to verity for yourself whether or not these functions are valid and provide the answer you seek. $\endgroup$
    – poetasis
    Apr 22, 2019 at 17:38

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


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