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?
 A: 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
A: Notation: All variable-symbols are positive integers.
Theorem 1. If $p$ is prime and $p\equiv 1 \pmod 4$ there exist $a,b$ with $a^2+b^2=p.$ Note that we must have  $\gcd(a,b)=1.$
Corollary. If $p$ is prime and $p\equiv 1 \pmod 4$,  there exist $x,y$ with $p^2=x^2+y^2$ and $\gcd (x,y)=1.$ Proof: By Theorem 1, let $p=a^2+b^2$ . Let $x=|a^2-b^2|$ and $y=2ab.$ We have $a\ne b$ (else $p=2a^2$ is not prime), so $x\ne 0.$  We have $p^2=x^2+y^2.$
Now $p^2$ is odd so $|a^2-b^2|$ is odd so $\gcd (|a^2-b^2|,2ab)=\gcd(|a^2-b^2|,ab).$ And $\gcd (a,b)=1,$ so if $q$ is prime and $q|ab$  then $$[(q|a\land q\not | b)\lor (q\not |a\land q|b) ],$$ which implies $q\not |\;(a^2-b^2).$ 
Theorem 2. If $q$ is prime and $q\equiv 3 \pmod 4$ then $q$ does not divide any $a^2+1.$ 
Theorem 3. There is no largest prime congruent to $1 \pmod 4.$ Proof: If $S$ is a finite non-empty set of primes, each congruent to $1\pmod 4$ then  any prime divisor of $(2\prod_{p\in S}p)^2+1$ does not belong to $S$, and is congruent to $1 \pmod 4,$ by Theorem 2. 
Lemma. If $x_1^2=a_1^2+b_1^2$ and $x_2^2=a_2^2+b_2^2$ where $1=\gcd (x_1,x_2)=\gcd (a_1,b_1)=\gcd (a_2,b_2)$ then $\gcd (a_1a_2+b_1b_2, a_1b_2-a_2b_1)=1.$
Proof: By contradiction, suppose prime $q$ divides both  $a_1a_2+b_1b_2$ and $a_1b_2-b_1a_2.$ Then modulo $q$ we have $$a_1a_2b_2\equiv -b_1b_2^2 \text { and }\;a_1b_2a_2\equiv b_1a_2^2$$  $$\text {implying }\; b_1(a+2^2+b_2^2)\equiv 0.$$  $$\text {Now }\;  q|b_1\implies q|(a_1ca+2+b_1b_2)-b_1b_2=a_1a_2\implies q|a_2$$ (because if $q|b_1$ and $\gcd (a_1,b_1)=1$ and  $q|a_1a_2$ then  $q|a_2$).  Similarly $$q|b_1\implies q|(a_1b_2-b_1a_2)+b_1a_2=a_1b_2\implies q|b_2.$$
So $q|b_1\implies \gcd (a_2,b_2)\geq q>1,$ which is false. (So far so good, but this is not the desired contradiction.) So from $b_1(a_2^2+b_2^2)\equiv 0\pmod q$ we conclude that $$q|(a_2^2+b_2^2)=x_2^2.$$ 
But by interchanging the subscripts 1,2 we also conclude that $$q|(a_1^2+b_1^2)=x_1^2.$$ 
This is a contradiction because $q$ is prime and $\gcd(x_1^2,x_2^2)=1. $ QED.
Putting this all together:
Let $p_1,..., p_{n+1}$ be $n+1$ distinct primes, each $\equiv 1 \pmod 4$.
Let $p_{n+1}^2=a_2^2+b_2^2$ with $\gcd (a_2,b_2)=1.$
If $\;\prod_{1=1}^np_i^2=a_1^2+b_1^2$ with $\gcd (a_1,b_1)=1,$ then $$\prod_{i=1}^{n+1}p_i^2=c^2+d^2$$ where $c=a_1a_2+b_1b_2$ and $d=|a_1b_2-a_2b_1|.$ And by the lemma we have $\gcd (c,d)=1.$  
