This question stems from another one that I asked earlier here. However, I have found a way to reformulate it and I believe this reformulation is sufficient to warrant posting it as a separate question. My initial question was, does there exist three different primitive Pythagorean triples such that,


Multiplying by two this can be rewritten as,


Looking at an individual term and applying the context of right triangles we can now obtain,


Now revisiting the original question and doing some algebra,



Where $\theta$ is the angle for the origin of the associated primitive Pythagorean triple. My belief (for whatever that's worth) is that for all primitive Pythagorean triples,


So far I have tested this for many triples but have not been able to find a counter example. As far as proving the latter I am pretty lost as far as finding an approach. For clarity I am asking, prove that there does not exist three primitive triples such that $$\sin{2\theta_1}+\sin{2\theta_2}=\sin{2\theta_3}$$ Or find a counter example. If you look at my initial question there was some helpful input and I would encourage you to look at it for some insight/context.

EDIT: I am amending this to be also including non primitive triples too. This makes sense because the solution to that should imply the other as we are dealing with angles and a normal triple should have the same angle as the scaled non primitive triple.

Edit2: Additionally, a,b,c can have no common factors.


1 Answer 1


Some possibilities:

The triangles can be normalized so that each $c_i = 1$.

You can then write each $a_i = m_i^2-n_i^2, b_i = 2m_in_i$ where the $m_i, n_i$ are rationals such that $m_i^2+n_i^2 = 1$.

Then $\sum_{i=1}^2m_in_i(m_i^2-n_i^2) =m_3n_3(m_3^2-n_3^2) $.

That's all that occurs to me right now.

  • $\begingroup$ Thanks for the insight! I will look into this to the best of my ability. $\endgroup$
    – PMaynard
    Dec 2, 2019 at 4:13
  • $\begingroup$ Do you have any more general thoughts on the question? I'm really unsure about the difficulty of this question (if it is worth pursuing) $\endgroup$
    – PMaynard
    Dec 2, 2019 at 4:16
  • $\begingroup$ Sorry, nothing more. It's just my feeling that expressing this in terms of rationals or integers would be more likely to lead to a solution than using the angles. $\endgroup$ Dec 2, 2019 at 4:18
  • $\begingroup$ Yeah thanks I think you might be right if you see the linked question this was the original approach but it didn't really go anywhere. $\endgroup$
    – PMaynard
    Dec 2, 2019 at 4:21

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