As we know that $x^2+y^2+z^2=8n+3,(n\in \mathbb N)\tag{1}$ has integer solutions $x,y,z\in \mathbb N.$
If $k\in \mathbb N$ has at least one prime factor which is $\equiv 3 \mod 4,$ then we call $k$ a good number.
If $x^2+y^2+z^2=8n+3\implies x,y,z$ are all good numbers, then we call $n$ a nice number.
For example, $n=8$ is a nice number since $x^2+y^2+z^2=8n+3\implies \{x,y,z\}=\{7,3,3\}.$
$n=8, 17, 26, 35, 44, 53, 62, 71, 152, 176\cdots$ are all nice numbers.
Is it true that if $n$ is a nice number then $x^2+y^2+z^2=8n+3\implies 3\mid xyz$?
PS: It seems that if $n$ is a nice number then $n\equiv 2 \pmod 3,$ hence $8n+3\equiv 1 \pmod 3,$ hence two of $x,y,z \equiv 0\pmod 3,$ and the rest one $\equiv \pm1\pmod 3.$
Here are some eaxmples($n<2000$):
$\{n,\{x,y,z\}\}$
{8,{{3,3,7}}}
{17,{{3,3,11},{3,7,9}}}
{26,{{3,9,11},{7,9,9}}}
{35,{{3,7,15},{9,9,11}}}
{44,{{3,11,15},{7,9,15}}}
{53,{{9,11,15}}}
{62,{{3,7,21},{7,15,15}}}
{71,{{3,11,21},{7,9,21},{11,15,15}}}
{152,{{3,11,33},{7,9,33},{7,21,27}}}
{176,{{3,21,31},{15,15,31},{21,21,23}}}
{233,{{3,3,43},{7,27,33},{11,15,39}}}
{314,{{7,21,45},{9,15,47},{15,21,43}}}
{395,{{7,33,45},{11,21,51},{11,39,39},{15,27,47},{15,33,43}}}
{698,{{3,33,67},{9,45,59},{23,33,63},{31,45,51}}}
{743,{{3,3,77},{15,31,69},{21,45,59},{27,27,67},{39,45,49}}}
{887,{{3,23,81},{3,51,67},{9,33,77},{21,27,77},{23,51,63},{27,49,63},{33,39,67}}}
{1265,{{3,33,95},{3,67,75},{9,59,81},{15,63,77},{23,45,87},{23,63,75},{31,51,81}}}
{1328,{{3,3,103},{9,39,95},{27,63,77},{39,59,75},{49,51,75}}}
{1832,{{3,3,121},{3,31,117},{3,75,95},{9,63,103},{9,77,93},{21,23,117},{23,81,87},{39,67,93},{45,45,103},{51,67,87},{63,69,77}}}
{1958,{{3,23,123},{9,81,95},{23,87,87},{33,63,103},{33,77,93},{59,75,81}}}