The answer is yes. For every side C in a primitive, a side A can be found in another primitive. For example, $(3,4,5)\rightarrow (5,12,13)\rightarrow (13,84,85)$ and these can be joined to make a single 4-dimensional figure. Let's reason this backward. If we have $5,12,13$, we can replace $5^2$ with $3^2+4^2$ and not affect the equation. Moving forward, $\quad 3^3+4^2+12^2+84^2=85^2.\quad$ One formula (mine) for generating Pythagorean triples is:
$$A=(2n-1)^2+2(2n-1)k\qquad B=2(2n-1)k+2k^2\qquad C=(2n-1)^2+2(2n-1)k+2k^2$$
It creates sets within the subset of triples where $GCD(A,B,C)=(2m-1)^2, m\in\mathbb{N}$ as shown here
$$\begin{array}{c|c|c|c|c|c|c|}
n & k=1 & k=2 & k=3 & k=4 & k=5 & k=6 \\ \hline
Set_1 & 3,4,5 & 5,12,13& 7,24,25& 9,40,41& 11,60,61 & 13,84,85 \\ \hline
Set_2 & 15,8,17 & 21,20,29 &27,36,45 &33,56,65 & 39,80,89 & 45,108,117 \\ \hline
Set_3 & 35,12,37 & 45,28,53 &55,48,73 &65,72,97 & 75,100,125 & 85,132,157 \\ \hline
Set_{4} &63,16,65 &77,36,85 &91,60,109 &105,88,137 &119,120,169 & 133,156,205 \\ \hline
Set_{5} &99,20,101 &117,44,125 &135,72,153 &153,104,185 &171,140,221 & 189,180,261 \\ \hline
\end{array}$$
If you have any side C, you can find a match in $Set_1$––side-A in $Set_1$ contains all odd numbers greater than one and all members of $Set_1$ are primitive––and sometimes elsewhere. If we solve the $A$-function for $k$, we can find a matching $A$, if it exists, with defined finite search limits. Any of the $n$-values that yield an integer $k$ gives us a triple.
$$A=(2n-1)^2+2(2n-1)k\quad\longrightarrow\quad k_A=\frac{A-(2n-1)^2}{2(2n-1)}\quad\text{where}\quad 1\le n\le\biggl\lfloor\frac{\sqrt{A+1}}{2}\biggr\rfloor $$ for example, beginning with $\qquad F(2,4)=(33,56,65)\qquad$ we have
$$\\ A=65\quad \longrightarrow \quad 1\le n\le\bigg\lfloor\frac{\sqrt{65+1}}{2}\bigg\rfloor=4 \quad\text{and we find} \quad n \in \{1,3\}\quad \Rightarrow \quad k\in\{32,4\}\\ $$
$$F(1,32)=(65,2112,2113)\qquad F(3,4)=(65,72,97)$$
Here, we have already created a new tuple where
$$33^2+56^2+72^2=97^2$$
How about
$$(3,4,5)\&(5,12,13)\&(13,84,85)\&(85,132,157)\&(157,12324,12325)\\\Longrightarrow 3^2+4^2+12^2+84^2+132^2+12324^2=12325^2$$
It is easy to see that the process can continue to infinity.