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 $n$...
 A: 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.
A: "OP" required numerical solution's to the below equation:
$x_1^{n-1} + x_2^{n-1}+x_3^{n-1}............. + x_{n-1}^{n-1} = y^{n-1}$ --(1)
For, $n=6$ we get:
$x_1^{5} + x_2^{5}+x_3^{5}+x_4^{5}+ x_{5}^{5} = y^{5}$
example is:  $(19,43,46,47,67)^5=(72)^5$

For, $n=5$
$x_1^{4} + x_2^{4}+x_3^{5}+x_4^{4}= y^{4}$
example is:    $(30,120,272,315)^4=(353)^4$

For, $n=4$
$x_1^{3} + x_2^{3}+x_3^{3}= y^{3}$
example is:      $(3,4,5)^3=(6)^3$

For, $n=3$
$x_1^{2} + x_2^{2}= y^{4}$
example is:      $(3,4)^2=(5)^2$
Note: There are numerical solution's for equation (1) above, for 
degree seven & eight but none yet for degree six.
A: I think it is still an open problem, wheather (k -1 -n) has finite or infinite solutions. Where k is degree of the equation & n is number of terms on the RHS. Mathematician Titus pizzas wrote an article some ten year's ago. And he ask's the same question. His Webpage link is given below.
[ http://sites.google.com/site/tpiezas ]. Click on algebraic identities. Click on Ramanujan pages. Click on article #15, Euler's conjecture. 
