All $C$-values are of the form $4n-1.\quad$ Here is a simple  BASIC program that I used to generate the $C$-values (in my other answer) that have $4$ matching primitive triples each. Sometimes there will be more than $4$ but those extras will be non-primitive. There will always be $2^{n-1}$ primitive triples for a $C$-value where $n$ is the number or prime factors of $C$. For instance, $1105=5\times13\times17$ so there are $2^2$ primitives. The number of primitives will be $2^{n-1}$ where $n$ is the number of unique prime factors of $C$. After the "run", the next section shows which ones produce all primitives.

     100 print "input limit";
     110 input l1
     120 for n1 = 1 to l1
     130     c1=4*n1+1
     140     m1=int((1+sqrt(2*c1-1))/2)
     150     m2=int(sqrt(c1-1))
     160     c9=0
     170     for m0=m1 to m2
     180         k0=sqrt(c1-m0^2)
     190         if k0=int(k0)
     200            c9=c9+1
     210         endif
     220      next m0
     230      if c9=4
     240         print c1,
     250      endif
     310 next n1


Here is a sample run where $l1=3000$

1105 	1625 	1885 	2125 	2405 	2465 	2665 	3145 	3445 	3485 	3625 	3965 	4225 	4505 	4625 	4745 	5125 	5185 	5365 	5785 	5945 	6205 	6305 	6409 	6565 	6625 	7085 	7225 	7345 	7565 	7585 	7625 	7685 	8177 	8245 	8585 	8845 	8905 	9061 	9125 	9265 	9605 	9685 	9805 	9945 	10205 	10585 	10865 	10985 	11125 	11245 	11285 	11645 	11713 	11765 

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Added: 

The following $C$-values have $4$ matching primitive Pythagorean triple each where

$$a_1^2+b_1^2=a_2^2+b_2^2=a_3^2+b_3^2=a_4^2+b_4^2=c^2$$

$$1105, 1885, 2405, 2465, 2665, 3145, 3445, 3485, 3965, 4505, 5185,  5365, 5785\\
 5945, 6205, 6305, 6409, 6565, 7085, 7345, 7565, 7585, 7685, 8177, 8245, 8585\\ 8845, 8905, 9061, 9565, 9605, 9685, 9805, 10205, 10585, 10865, 11245 11285\\
 11645, 11713, 11765, 12505, 12545,12665, 12805, 12905, 13345, 13481, 13505\\
13949, 14065, 14645, 14705, 14885, 14965, 15145, 15385, 15457, 15665, 15805$$

We can find these triple by solving Euclid's formula for $C$ and testing a defined range of m-values to see which, if any, yield integers.
$$ \quad A=m^2-k^2,\quad B=2mk,\quad C=m^2+k^2\quad$$

Here is an example using $C=64$ to find two primitive triples. 
$\qquad 1105$ would have yielded $4$

\begin{equation}
C=m^2+k^2\implies k=\sqrt{C-m^2}\\
\text{for}\qquad \bigg\lfloor\frac{ 1+\sqrt{2C-1}}{2}\bigg\rfloor \le m \le \lfloor\sqrt{C-1}\rfloor
\end{equation}
The lower limit ensures $m>k$ and the upper limit ensures $k\in\mathbb{N}$.
$$C=65\implies \bigg\lfloor\frac{ 1+\sqrt{130-1}}{2}\bigg\rfloor=6 \le m \le \lfloor\sqrt{65-1}\rfloor=8\\
\land \quad m\in\{7,8\}\implies k\in\{4,1\}\\$$
$$F(7,4)=(33,56,65)\qquad \qquad F(8,1)=(63,16,65) $$

If we were to use $C=1105$, we would find

$$f(24,23)=(47,1104,1105)\quad 
f(31,12)=(817,744,1105)\\ 
f(32,9)=(943,576,1105)\quad 
f(33,4)=(1073,264,1105)$$