If $3\mid a,b,c$ and $n=a^2+b^2+c^2$, prove that there exist $x,y,z$ such that $n=x^2+y^2+z^2$, where $3\nmid x,y,z$. Here $n\in\mathbb N$, $a,b,c,x,y,z\in\mathbb Z$.
This problem is originally stated as:
If a natural number $n$ can be expressed as a sum of $3$ squares of integers that are divisible by $3$, prove that it can be expressed as a sum of $3$ squares of integers that are not divisible by $3$ as well.
I'm not sure if $0$ is included in the set of natural numbers in this case but let's assume it is.
So I've checked divisibility by $9$ and it doesn't contradict the possibility. Divisibility by $3$ in this case guarantees the same remainders.
I've tried marking $a,b,c,x,y,z$ as $3k,3l,3m,3d\pm1,3e\pm1,3f\pm1$ respectively. This allows us to work with integers with no constraints. It brings me to this: $$9k^2+9l^2+9m^2=9(d^2+e^2+f^2)\pm6d\pm6e\pm6f+3=n$$
And now we can just prove that this equality $$9k^2+9l^2+9m^2=9(d^2+e^2+f^2)\pm6d\pm6e\pm6f+3$$
can always hold when all the variables are just integers.
Some ideas would be great. Thanks.