<s>[INST] Is the set $M = \{d(n) \mid n \in S\}$ unbounded? 
Consider the set $S = \{c \in \mathbb{Z}^+ \mid a^2+b^2 = c^2, \text{for some} \text{ } \text{coprime positive integers} \text{ } a,b\}$. Let $d(n)$ denote the number of positive integer divisors of a positive integer $n$. Is the set $M = \{d(n) \mid n \in S\}$ unbounded?

In other words, I am wondering if the hypotenuse of a primitive right triangle can have an arbitrarily large amount of divisors. How can we relate the divisor function to a primitive right triangle?
 [/INST] If you take $n = 5 \cdot 13 \cdot 17 \cdot \cdots \cdot p_r$ is the product of $r$ distinct primes, all $p \equiv 1 \pmod 4,$ then $n$ is the hypoteneuse of a primitive Pythagorean triangle. Also $d(n) = 2^r$   
Fermat: every prime $p  \equiv 1 \pmod 4$ can be written as $p = a^2 + b^2$
I can't seem to remember the full result on coprime stuff, so I'm going the cheap way here. From the Brahmagupta identity we have $n = A^2 + B^2.$ As $n$ is squarefree we know that $\gcd(A,B) = 1.$ As $n \equiv 1 \pmod 4$ we know that one of $A,B$ is even and the other odd. We then get
$$ (2AB)^2 + (A^2 - B^2)^2 = n^2. $$
We have $A^2 - B^2$ odd. If we have some prime  $q | \gcd(2AB, A^2 - B^2),$ we know $q$ is odd. We also know $q | n.$ more to come... Oh, I see, as $q$ is odd and $q | (A-B)(A+B),$ and $q$ divides one of $A,B,$ but $q$ also divides at least one of $A-B, A+B,$ it folows that $q$ actually divides both $A,B.$ This contradicts $\gcd(A,B) = 1.$ Therefore, $ \gcd(2AB, A^2 - B^2) = 1$ and
$$   2AB, A^2 - B^2,n $$
is primitive.
Brahmagupta's identity
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