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

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.

• 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. Jan 20 '18 at 7:59
• Sorry for the late reply. Thank you, sir! Jan 24 '18 at 4:18
• If you multiply any hypotenuse by a prime other than 2, the result will be odd. To find out whether or not there exists $1$-or-more triples with that hypotenuse, see matching sides of Pythagorean triples. May 27 '19 at 17:21

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

• You are welcome! Jan 20 '18 at 8:14

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

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 $$(2+i)^2·(2\pm3i)·(4\pm i)$$

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

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 $$\quad A=m^2+n^2\quad B=2mn\quad C=m^2+n^2$$.

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