How many variations of new primitive Pythagorean triples are there when the hypotenuse is multiplied by a prime?

Question

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.

Answer 1

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

Answer 2

Can anyone explain to me why these trends are here?

Let $n$ be a Natural number that factors as $\def\mod{\operatorname{mod}}$

$$ n=\!\!\prod_{p\text{ prime}} p^{m_p} $$

• If $m_p>0$ for some prime with $p\neq1\mod 4$, then there is no primitive Pythagorean triple $(a,b,n)$.

• Let $n$ be composed only of primes $p=1\mod 4$, and $m$ be the number of such primes (without multiplicity). Then there are exactly $2^{m-1}$ primitive Pythagorean triples of the form $$(a,b,n)\quad\text{ where }\quad 0<a<b $$

For example, $65=5·13$ gives rise to $2^{2-1}=2$ such triples, same for $85=5·17$. Thus, $c=5525=5^2·13·17$ gives rise to 4 primitive triples, namely:

• $(1036, 5427, c)$
• $(2044, 5133, c)$
• $(2163, 5084, c)$
• $(3124, 4557, c)$

To see / compute this, factor $c$ over $\Bbb Z[i]$, the Gaussian integers. $c$ will decompose into $2m$ prime factors (not counting multiplicity) of $m$ conjugate pairs. To get all triples, take a prime of the first$^1$ pair to it's appropriate power $m_p$, and multiply it with whatever conjugate of the remaining pairs to their's powers $m_p$.$\def\cc{\mathrm{c.c}}$

For example, up to units we have the factorizations $17=(4+i)_\cc$, $5=(2+i)_\cc$ and $13=(2+3i)_\cc$ where "c.c." means to multiply with the according complex conjugate. This gives – up to order, units and complex conjugation – the 4 distinct products of norm 5525 $$ (2+i)^2·(2\pm3i)·(4\pm i)$$

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$^1$The order does not matter, just fix one order at your preference.

Answer 3

Too long for a comment:

@emacs drives me nuts You pointed out something I never thought about before: that the number of primitive triples for a given $C$ is $2^{X-1}$ where $x$ is the number of prime factors of $C$ that take the form $4x+1, x\in\mathbb{N}$. For example:$\quad 1185665 =5*13*17*29*37$ and has $16$ primitive triples. Note: $f(m,n)$ refers to Euclid's formula where $\quad A=m^2+n^2\quad B=2mn\quad C=m^2+n^2$.

$f(796,743)=(81567,1182856,1185665)\\ f(856,673)=(279807,1152176,1185665)\\ f(863,664)=(303873,1146064,1185665)\\ f(904,607)=(448767,1097456,1185665)\\ f(908,601)=(463263,1091416,1185665)\\ f(929,568)=(540417,1055344,1185665)\\ f(961,512)=(661377,984064,1185665)\\ f(992,449)=(782463,890816,1185665)\\ f(1028,359)=(927903,738104,1185665)\\ f(1049,292)=(1015137,612616,1185665)\\ f(1052,281)=(1027743,591224,1185665)\\ f(1063,236)=(1074273,501736,1185665)\\ f(1072,191)=(1112703,409504,1185665)\\ f(1076,167)=(1129887,359384,1185665)\\ f(1084,103)=(1164447,223304,1185665)\\ f(1087,64)=(1177473,139136,1185665)\\$