Square equal to sum of three squares For which integers $n$ there exists integers $0\le a,b,c < n$ such that $n^2=a^2+b^2+c^2$?
I made the following observations:

*

*For $n=1$ and $n=0$ those integers doesn't exist.


*If $n$ is a power of 2 those integers doesn't exist. Let $n=2^m$ with $m>0$ the smallest power of 2 for which there exists $a,b,c$ such that $\left (2^m\right )^2=4^m=a^2+b^2+c^2$. Since $4^m$ is divisible by 4, $a^2+b^2+c^2$ has to be divisible by 4 too. This is only possible if $a^2\equiv b^2\equiv c^2\equiv 0\pmod 4$, so we can write $a=2a',b=2b',c=2c'$ with $a',b',c'\in \mathbb{N}$. But then we get $\left (2^{m-1}\right )^2=4^{m-1}=a'^2+b'^2+c'^2$, so $m=1$, otherwise $2^m$ wouldn't be the smallest power of two with this property. It is easy to check that $n=2$ doesn't work, so for $n=2^m$ the statement doesn't hold.


*I suspect (but can't prove) that for all other values the statement holds. It would be enough to prove that for all odd primes $p$ there exists $a,b,c$ such that $p^2=a^2+b^2+c^2$, since for all other values of $n$ there exist some $p,m$ such that $n=pm$. Then we get $n^2=(pm)^2=(ma)^2+(mb)^2+(mc)^2$.

You are correct: If $p > 2$ is prime, then $p^2$ can always be written as the sum of three squares at least two of which are non-zero.
Let $s(n)$ denote the number of ways of writing $n = a^2 + b^2 + c^2$, where $a$, $b$, and $c$ are integers (positive or negative) and not accounting for symmetries. One has $s(1) = 6$.
If $p > 2$ is prime, then $p^2$ can be written as a sum of three squares (including degenerate examples) in
$$6\left(p + 1 - \left( \frac{-1}{p} \right)\right)$$
ways. (For a reference, see https://mathoverflow.net/questions/3596/is-there-a-simple-way-to-compute-the-number-of-ways-to-write-a-positive-integer). For example, if $p = 3$, then $(-1/3) = -1$ so we get $30$ ways, and indeed
$$3^2 = (\pm 3)^2 + 0^2 + 0^2 = 0^2 + (\pm 3)^2 + 0^2  = 0^2 + 0^2 + (\pm 3)^2,$$
giving $3 \times 2 = 6$ ways, and
$$3^2 = (\pm 2)^2 + (\pm 2)^2 + (\pm 1)^2 = (\pm 2)^2 + (\pm 1)^2 + (\pm 2)^2 = (\pm 1)^2 + (\pm 2)^2 + (\pm 2)^2$$
giving $3 \times 8 = 24$ ways. The examples you wish to rule out at the ones where either $a$, $b$, or $c$ is $\pm p$, and this gives $6$ solutions. Hence you simply need to observe that $p + 1 - (-1/p) > 1$, which is always true.
