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When I was looking at all the Pythagorean triples from 1 to 1000, I noticed a certain trend which I cannot seem to explain. Given $a, b, c$ are integers where $c$ is the hypotenuse, and $a < b$, the trends were

  1. For all $(a, b, c)$s where $c$ is a prime, there is only one pair possible. Here, not all primes have this property. There are primes where there are no triplets corresponding to it yet I could not find a consistent property between these primes.
  2. When c is multiplied by another prime that has a property of having a triplet, if that prime is a different prime than the first prime, there will be two new primitive Pythagorean pairs formed. For example, for (3, 4, 5) and (5, 12, 13), when we look at c = 65 which is 5 times 13, there will be 4 pairs which are (52, 39), (56, 33), (60, 25), (63, 16) where (65, 56, 33) and (65, 63, 16) are new primitive Pythagorean triples.
  3. If c is multiplied by the same prime, c, there will only be 1 new primitive Pythagorean triple produced.

Can anyone explain to me why these trends are here? Thanks in advance! Below I will document my failed approach just in case it will help clarify my question.

My(unsuccessful) approach

For 1, I tried to use a proof by contradiction where I tried to prove that $$c^2 = a_1^2 + b_1^2 = a_2^2 + b_2^2$$ was impossible. Yet sadly I could not find any contradiction.(I assume there is one though)

For 2, where $c_0, c_1$ are primes, when looking at $c_0c_1$, $$c_0^2c_1^2 = (a_0^2 + b_0^2)(a_1^2 + b_1^2) = a_0^2c_1^2 + b_0^2c_1^2 = a_1^2c_0^2 + b_1^2c_0^2$$ where the other two triples which will be primitive will come from the expansion of it.

For 3, the basic approach was the same as 2 except for the fact that $c_0 = c_1$ I failed to find any primitive triplets.

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  • $\begingroup$ Look at Fermat's theorem on the sum of two squares. Also, the sum of two squares times the sum of two squares is again the sum of two squares. $\endgroup$ – saulspatz Jan 20 '18 at 7:59
  • $\begingroup$ Sorry for the late reply. Thank you, sir! $\endgroup$ – Isamu Isozaki Jan 24 '18 at 4:18
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For the first statement.

If $c=4n+1$, where $n$ is natural and $c$ is a prime number then it's true.

Also, we have $$c_0^2c_1^2=(a_0^2+b_0^2)(a_1^2+b_1^2)=a_0^2a_1^2+a_0^2b_1^2+b_0^2a_1^2+b_0^2b_1^2=$$ $$=(a_0a_1+b_0b_1)^2+(a_0b_1-a_1b_0)^2=(a_0a_1-b_0b_1)^2+(a_0b_1+a_1b_0)^2.$$

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  • $\begingroup$ You are welcome! $\endgroup$ – Michael Rozenberg Jan 20 '18 at 8:14

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