TITLE: Intersecting Sets of Pythagorean Triples with Common Hypotenuses
QUESTION [6 upvotes]: For any $r\in\mathbb{N}$, let $A_r$ denote the set of all natural numbers that are potentially a side of a Pythagorean triple with hypotenuse $r$.
Given any $N\in\mathbb{N}$, does there exist $r,s$ such that $A_r\cap A_s$ contains at least $N$ elements?
I've been looking at the parameterization of Pythagorean triples as $k(m^2-n^2,2mn,m^2+n^2)$, and considering numbers $r$ that can be written as the sum of two squares in a large number of ways.  But the components of this sum seem rather unpredictable, and I'm unsure how to proceed in this fashion.

REPLY [2 votes]: Take integers that are sums of two squares; this set is multiplicatively closed because of the identity $(a^2+b^2)(c^2+d^2) = (ad-bc)^2 + (ac+bd)^2$
Now interchanging $a$ and $b$, the new  rhs gives another expression for the same  LHS as sum of two squares. Now this explodes when a number is a product of many terms which are individually sum of two squares. And primes 1 mod 4 are by Fermat sum of two squares, and a number that is a product of many such primes will provide examples you need.