<s>[INST] Are all integers not congruent to 6 modulo 9 of the form $x^3+y^3+3z^2$? Are all integers not congruent to 6 modulo 9 of the form $x^3+y^3+3z^2$
for possibly negative integers $x,y,z$?
We have the identity $ (-t)^3+(t-1)^3+3 t^2=3t-1$.
The only congruence obstruction we found is 6 modulo 9.
 [/INST] We can also say the each $n \in \mathbb{Z}$ with $n \equiv 3 \pmod 9$ is representable as $x^3 + y^3 + 3z^2$. This is because
$$(-t)^3 + (t-9)^3 + 3(3t -13)^2 = 9t - 222$$
and $-222 \equiv 3 \pmod 9$.
So, along with the identity in the question we can represent each integer congruent to $2$, $3$, $5$, or $8 \pmod 9$. This leaves only $0$, $1$, $4$, and $7 \pmod 9$.
More generally we have
$$(-t)^3 + (t - a^2)^3 + 3(at - b)^2 = 3a(a^3 - 2b)t + 3b^2 - a^6$$
which is congruent to $-a^6 \pmod 3$. This is either $0$ or $-1 \pmod 3$ depending on if $3$ divides $a$ or not. Taking $a = 3$ and $b = 13$ we obtain the first equation which takes care of $n \equiv 3 \pmod 9$. If we take $a = 3$ and $b = 12$ we get $27t - 297$.
This gives us $n \equiv 0 \equiv -297 \pmod{27}$ which is some further progress. Hence, $0 \pmod 9$ can be resolved by finding $9 \pmod {27}$ and $18 \pmod{27}$.
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