Let $n \ge 2$ be a positive integer. Do there exist $n$ non-zero distinct integers such that the sum of their square is a perfect square and their product is a nth power?
For $n=2$ the answer is no, by infinite descent. Is this true for all $n$? What happens for other values of $n$?
There are such solutions for each $n>2$.
We seek distinct nonzero integers $x_1,\ldots,x_n$ such that $\sum_{i=1}^n x_i^2 = y^2$ and $\prod_{i=1}^n x_i = z^n$. These equations are homogeneous, so it is enough to consider the distinct nonzero rationals $r_i := x_i / z$, which satisfy $\sum_{i=1}^n r_i^2 = (y/z)^2$ and $\prod_{i=1}^b r_i = 1$: given distinct nonzero rationals $r_1,\ldots,r_n$ with $\prod_{i=1}^n r_i = 1$ such that $\sum_{i=1}^n r_i^2 = \eta^2$ for some rational $\eta$, we can clear common factors to recover an integer solution $x_1,\ldots,x_n$.
Now if we fix $n-2$ of the $r_i$, say $r_3,\ldots,r_n$, then the remaining two (here $r_1,r_2$) together with $\eta$ satisfy the equations $$ r_1 r_2 = \prod_{i=3}^n r_i^{-1}, \quad r_1^2 + r_2^2 + \sum_{i=3}^n r_i^2 = \eta^2, $$ which give an elliptic curve. So, once we have a single "random" solution $(r_1,\ldots,r_n,\eta)$, we get a pair $(r_1,r_2,\pm\eta)$ of rational points on that elliptic curve, the group law to get infinitely many others with the same $r_3,\ldots,r_n$; we can then start from any of those points, switch to another elliptic curve and use its group law, etc. until we get a "Zariski-dense" set of rational points on the variety $\prod_{i=1}^n r_i = 1, \sum_{i=1}^n r_i^2 = \eta^2$. In particular, we'll find solutions with distinct $r_i$. even if our initial solution had some coincidences. For example, if $n=4$ we can start from $(1/2,1,1,2;5/2)$, fix $(r_3,r_4)=(1,2)$, and solve for $r_2$ to get $r_2 = 1/(2r_1)$, getting the elliptic curve $4r_1^4 + 20 r_1^2 + 1 = s^2$ (where $s = 2r_1\eta$). The osculating parabola at $(r,s) = (1/2,5/2)$ is $s = (166 r_1^2 + 384 r_1 + 79) / 125$, giving the new solution $r_1 = 391/182$ and then $(r_1,r_2,r_3,r_4;\eta) = (391/182, 91/391, 1, 2; 221285/71162)$ which scales to the integer solution $(152881, 16562, 71162, 142324; 221285, 71162)$ with distinct variables.
So it remains to find a starting solution for each $n>2$. There are already examples for $3 \leq n \leq 9$ in the comments:
n r1, ..., rn; eta
3 1/3, 3/4, 4; 49/12
4 1/10, 5/4, 5/2, 16/5; 16/5
5 1/4, 1/3, 1, 2, 6; 77/12
6 1/2, 2/3, 5/6, 9/10, 3/2, 8/3; 17/5
7 1/6, 2/3, 3/4, 4/3, 3/2, 2, 3; 17/4 [W. Jagy]
8 1/6, 5/6, 9/10, 1, 6/5, 4/3, 2, 5/2; 4
9 1/6, 1/5, 4/15, 1/3, 1, 5/2, 4, 9/2, 15/2; 10
To go further, we search for rational $r_1,\ldots,r_7$, not necessarily distinct, such that $\prod_{i=1}^r r_i = 1$ and $\rho := \sum_{i=1}^r r_i^2$ is an integer (not necessarily a square); $7$ was the first to easily give plentiful choices:
rho r1, ..., r7
--- ----------------
10 1/3, 5/6, 5/6, 6/5, 3/2, 3/2, 8/5
12 1/6, 5/6, 6/5, 3/2, 3/2, 8/5, 5/3
13 1/6, 2/3, 4/3, 3/2, 3/2, 3/2, 2
15 1/6, 5/6, 9/10, 6/5, 4/3, 2, 5/2
16 1/10, 5/6, 4/3, 4/3, 3/2, 9/5, 5/2
17 1/6, 2/3, 9/10, 6/5, 4/3, 5/2, 5/2
18 1/6, 1/3, 3/2, 3/2, 3/2, 2, 8/3
19 4/15, 1/2, 2/3, 5/6, 2, 5/2, 27/10
20 3/10, 2/5, 5/6, 5/6, 3/2, 8/3, 3
21 4/15, 1/2, 2/3, 2/3, 5/2, 5/2, 27/10
22 1/6, 1/3, 9/10, 6/5, 5/2, 5/2, 8/3
We can now mix and match these $r_1,\ldots,r_7$ together with some $r_i=1$ to find an initial solution for each $n>9$. For $n=10$, take the $\rho=22$ solution with $r_8=r_9=r_{10}=1$ to make $\sum_{i=1}^{10} r_i^2 = 25$; likewise for $n=11,12,13,\ldots,17$ with $\rho = 25 + 7 - n$; then for $n=18$ use the $\rho=10$ and $\rho=22$ sets with $r_{15}=r_{16}=r_{17}=r_{18}=1$ to make $\sum_{i=1}^{10} r_i^2 = 36$; for the next few $n$, change $\rho=22$ to $\rho=21,20,19,\ldots$; as $n$ increases further, move up to $\sum_{i=1}^n r_i^2 = 49$, $64$, $81$, etc. There are more than enough options to get a starting solution for each $n$, and then the elliptic-curve technique does the rest.
