Are there infinitely many primitive Pythagorean $n$-tuples? Let's define a Pythagorean $n$-tuple ($n \geq 3$) as a tuple of distinct natural numbers $(x_1, ... , x_{n-1}, y)$, such that $x_1^{n-1} + ... + x_{n-1}^{n-1} = y^{n-1}$. Let's call a Pythagorean $n$-tuple primitive if $GCD(x_1, ... , x_{n-1}, y) = 1$

It is a rather well known fact, that there are infinitely many primitive Pythagorean triples. Indeed, $(2k(k + 1))^2 + (2k + 1)^2 = (2k(k + 1) + 1)^2$ $\forall k \in \mathbb{N}$.

But is it true that $\forall n \geq 3$ there are infinitely many primitive Pythagorean $n$-tuples?

No construction, similar to the one used for $n = 3$ comes to my mind for arbitrary $n$...
For $n=4$ we have this [1] infinite parametrised set of solutions for 1:
$$(9t^4)^3 + (-9t^4 \mp 3t)^3 + (\pm 9t^3 + 1)^3 = 1$$

[1] Kurt Mahler, Note On Hypothesis K of Hardy and Littlewood, Journal of the London Mathematical Society 11 (1936), 136–138.
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
$$\begin{array}{c|c|c|c|c|c|c|} 
n &amp; k=1 &amp; k=2 &amp; k=3 &amp; k=4 &amp; k=5   &amp;  k=6  \\ \hline
Set_1 &amp; 3,4,5 &amp; 5,12,13&amp; 7,24,25&amp; 9,40,41&amp; 11,60,61 &amp; 13,84,85 \\ \hline
Set_2 &amp; 15,8,17 &amp; 21,20,29 &amp;27,36,45 &amp;33,56,65 &amp; 39,80,89 &amp; 45,108,117 \\ \hline
Set_3 &amp; 35,12,37 &amp; 45,28,53 &amp;55,48,73 &amp;65,72,97 &amp;  75,100,125 &amp; 85,132,157 \\ \hline
Set_{4} &amp;63,16,65 &amp;77,36,85 &amp;91,60,109 &amp;105,88,137 &amp;119,120,169 &amp; 133,156,205 \\ \hline
Set_{5} &amp;99,20,101 &amp;117,44,125 &amp;135,72,153 &amp;153,104,185 &amp;171,140,221 &amp; 189,180,261 \\ \hline
\end{array}$$

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 
$$(3,4,5)\&amp;(5,12,13)\&amp;(13,84,85)\&amp;(85,132,157)\&amp;(157,12324,12325)\\\Longrightarrow 3^2+4^2+12^2+84^2+132^2+12324^2=12325^2$$

It is easy to see that the process can continue to infinity.