Is there an upper bound on the number of primtive pythagorean triples in which some odd number can be a non-hypoteneuse edge?

Take $15$ for example, we have $15,8,17$ then $15,112,113$ but it's not clear we can keep going forever.

I suspect if infinitely many odd integers were the non-hypoteneuse edge of infinitely many primitive pythagorean triples then we might use the diagonal argument to imply uncountably many different pythagorean triples.

But that would contradict that for $a^2+b^2=c^2$ we have $\left(\dfrac{a}{c}\right)^2+\left(\dfrac{b}{c}\right)^2=1$, putting every triple into bijection with a set of rational points on the unit circle and making them countable. So by that argument there are at most finitely many which are members of infinitely many triples.

  • $\begingroup$ Say the number of interest is $a$. Every Pythagorean triple in which $a$ occurs as a non-hypotenuse corresponds to a factorisation $a^2 = (c-b)(c+b)$. What condition says the Pythagorean triple is primitive? $\endgroup$ – Daniel Fischer Apr 5 '18 at 19:36
  • $\begingroup$ I meant, "what condition for the factorisation $(c-b)(c+b)$". $\endgroup$ – Daniel Fischer Apr 5 '18 at 19:44
  • $\begingroup$ @DanielFischer I've no idea really. I can see if it wasn't primitive then $(c-b), (c+b)$ would not be coprime but I can't show the contrapositive. $\endgroup$ – user334732 Apr 5 '18 at 20:29
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    $\begingroup$ A common divisor of $c-b$ and $c+b$ also divides $(c - b) + (c + b) = 2c$ and $(c + b) - (c - b) = 2b$. If $a$ is odd, then $c - b$ and $c + b$ are both odd, so $\gcd(c-b,c+b) = \gcd(b,c)$. So for odd $a$, the primitive Pythagorean triples containing $a$ as a leg are in bijection with the factorisations of $a^2$ into two coprime factors (where order of the factors doesn't count). Given an arbitrary $n$, can you find an $a$ that admits at least $n$ factorisations of $a^2$ into two coprime factors? $\endgroup$ – Daniel Fischer Apr 5 '18 at 20:36
  • $\begingroup$ @DanielFischer I think if $f(p,2)$ counts the ways to partition $p$ elements into two groups then we would require $p$ the number of distinct prime factors in $a$ to be sufficient that $n\leq f(p,2)$. Is that right? $\endgroup$ – user334732 Apr 5 '18 at 21:08

A simple way to show there is not is to use the parameterization of primitive Pythagorean triples. Let $m,n$ be coprime and of opposite parity. Then $m^2-n^2,2mn,m^2+n^2$ is a primitive Pythagorean triple. For our use, let $m$ be even and $n=1$, giving the fact that $m^2-1,2m,m^2+1$ is a Pythagorean triple. If you want an odd number that is a leg of $k$ Pythagorean triples we start with $$2^2-1,2\cdot 2,2^2+1\\4^2-1,2\cdot 4,4^2+1\\6^2-1,2\cdot 6,6^2+1\\ \ldots \\(2k)^2-1,2\cdot 2k,(2k)^2+1\\$$ Now multiply each by $(2^2-1)(4^2-1)(6^2-1)\ldots ((2k)^2-1)$ divided by its first element and we will have $k$ triangles all with odd leg $(2^2-1)(4^2-1)(6^2-1)\ldots ((2k)^2-1)$

  • $\begingroup$ thanks Ross, nice. If I understand correctly Daniel Fischer has also shown with with a bit of combinatorics it looks like we can actually derive a function for the number of triples in which any odd integer appears as a leg, based on its number of distinct prime factors. $\endgroup$ – user334732 Apr 6 '18 at 16:20
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    $\begingroup$ @ProducerofBS: that is correct. I thought this was a simpler approach to answer the question by actually constructing a number that it part of at least $k$ different triangles for any given $k$. In fact it will be part of many more for any reasonably large $k$. $\endgroup$ – Ross Millikan Apr 6 '18 at 16:24
  • $\begingroup$ You have answered the question so I feel bad for not accepting just yet but it would also be lovely to arrive at the formula Daniel has alluded to. I'll set aside some time to tackle it myself but in truth I judge myself unlikely to arrive at a reliable answer. $\endgroup$ – user334732 Apr 6 '18 at 16:26
  • $\begingroup$ @ProducerofBS: Daniel Fischer is talking about primitive pairs as well. If $a$ has $p$ distinct prime factors the number of factorizations of the type he says is $2^{p-1}$. All the powers of one prime have to go into the same factor to have the two factors coprime. If you just take all the subsets of the set of prime factors as one factor, you count every division twice, once when you pick a set and once when you pick its complement. $\endgroup$ – Ross Millikan Apr 20 '18 at 15:30
  • $\begingroup$ Thanks. I had come to the conclusion it would be $2^{p-1}$ in fact asked exactly that here: math.stackexchange.com/questions/2744881 as I wasn't totally confident in my own abilities. Happy for you to confirm that in either this answer or an answer to that. $\endgroup$ – user334732 Apr 20 '18 at 15:37

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