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.



Intersecting Sets of Pythagorean Triples with Common Hypotenuses
Yes, given any $N \in \mathbb{N}$ there do exist $r$ and $s$ so that $A\_{r} \cap A\_{s}$ contains at least $N$ elements. In fact, I will show that one can take $s = 6r$. (The approach I suggest below is definitely not the most efficient, nor is it along the lines you were pursuing.)


We want to find values of $r$ so that the equations
$$
C : a^2 = r^2 - b^2, \quad a^2 = 36r^2 - c^2
$$
have at least $N$ integer solutions for $a$. This system of two quadratic equations is isomorphic to the elliptic curve
$$ 
E : y^2 = x^3 - x^2 - 420x - 3168,
$$
which has infinitely many rational solutions. A generator for the group of rational points is $P = (69,540)$, and this generator maps to the solution $145^2 = 144^2 + 17^2$, $(6 \cdot 145)^2 = 144^2 + 858^2$, showing that $A\_{145}$ and $A\_{870}$ have (at least) one element in common. The point $2P$ is $(14113/576,352495/13824)$ and this
shows that $A\_{430106689}$ and $A\_{2580640134}$ also have an element in common (namely $101518560$). Taking the least common multiple of $145$ and $430106689$ shows that for $r = 145 \cdot 430106689$ there are at least two elements of $A\_{r} \cap A\_{6r}$.


Using the points $P$, $2P$, $\ldots$, $NP$, finding their images on $C$, and taking the least common multiple of the corresponding values of $n$ will give a value of $r$ so that $A\_{r} \cap A\_{6r}$ will have size at least $2N$. 


