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
- 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.
- 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.
- 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.
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.