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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.

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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 $2N$.

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  • $\begingroup$ I had a suspicions that Eliptic curves would be involved. Unfortunately, I havent studied them since My early undergraduate years, so your comment is a bit out of my area. But it seems entirely reasonable and correct, so I'll do some reading on this subject. Thank you very much! (I will vote once I am able, when my account is officially confirmed) $\endgroup$ – G. Flowers Mar 13 '14 at 4:14
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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.

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