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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 $n$?

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    $\begingroup$ There are solutions at least for $n=3,4,5$: for $n=3$, we have $$ (4,9,48;49,12), (3,36,128; 133,24), (56,72,147; 173,84) $$ and no other primitive examples up to $200$; for $n=4$, $$ (2,25,50,64;85,20), (4,30,75,90;121,30), (6,20,75,90;119,30), (10,30,36,75;89,30), (15,18,30,100;107,30) $$ and no others through $100$; and for $n=5$, $$ (3,4,12,24,72;77,12), (3,24,36,48,64;91,24) $$ and no others through $100$. $\endgroup$ – Noam D. Elkies Jan 30 at 15:12
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    $\begingroup$ . . . and for $n=6$ the first solution is $(15, 20, 25, 27, 45, 80; 102, 30)$ (it took about 3 minutes to try all sextuples up to $81$). $\endgroup$ – Noam D. Elkies Jan 30 at 15:20
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    $\begingroup$ n=7 36 24 18 16 9 8 2 sumsquares 3^2 17^2 prod 2^14 3^7 $\endgroup$ – Will Jagy Jan 30 at 19:41
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    $\begingroup$ n=8: (5, 25, 27, 30, 36, 40, 60, 75; 120, 30) $$ $$ There are solutions for each $n>2$; I should be able to post an answer later today if nobody else does it first. $\endgroup$ – Noam D. Elkies Jan 31 at 17:13
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    $\begingroup$ meanwhile, for n=9: $(5, 6, 8, 10, 30, 75, 120, 135, 225; 300, 30)$ $\endgroup$ – Noam D. Elkies Jan 31 at 20:56
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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.

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  • $\begingroup$ Thanks, Noam. .. $\endgroup$ – Will Jagy Feb 1 at 22:07
  • $\begingroup$ @NoamD.Elkies Thank you. $\endgroup$ – jack Feb 2 at 21:03

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