In thinking about this question sum of 3 squares that has now been marked as a duplicate
"If the numbers $(8k+7)$, $1$, $2$, $5$, $10$, $13$, $25$, $37$, $58$, $85$ and $130$ are all the primitives that that cannot be written as the sum of $3$ [non-zero squares], then the full list of numbers that cannot be written as the sum of three [non-zero squares] squares is $(8k+7)\times4^m$, $1\times4^m$, $2\times4^m$, $5\times4^m$, $10\times4^m$, $13\times4^m$, $25\times4^m$, $37\times4^m$, $58\times4^m$, $85\times4^m$ and $130\times4^m$ "
Does this second list represent the entire set of numbers that cannot be written as the sum of three non-zero squares?
If this is not the entire set can it at least be shown that the number of primitives not of the form $(8k+7)$ is finite.
[Note: I have not proved that $4^m$ is the only such multiplier of primitives so the above statement in quotes is a conjecture.]