<s>[INST] Hilbert's Identity for solving Waring's Problem Showing that $g(4)$ exist is quite easy using this identity
\begin{eqnarray}
   6(a^2 + b^2 + c^2 + d^2)^2 &=& (a + b)^4 + (a - b)^4 + (c + d)^4
     \nonumber
  \\  & &+(c-d)^4+(a+c)^4+(a-c)^4
  \nonumber
  \\ &  &+(b+d)^4+(b-d)^4+(a+d)^4
    \nonumber
  \\& &+(a-d)^4+(b+c)^4+(b-c)^4
  \nonumber
\end{eqnarray}
For $g(6)$:
$$60(a^2_1+a^2_2+a_3^2+a_4^2)^3=\sum_{i>j>k}(a_i\pm a_j\pm a_k)^6+2\sum_{i>j}(a_i\pm a_j)+36\sum_i a_i^6$$
And for $g(8)$
$$
5040(a_1^2+a_2^2+a_3^2+a_4^2)^4=6\sum(a_1\pm a_2\pm a_3\pm a_4)^8+\sum_{i>j>k}(2a_i\pm a_j\pm a_k)^8+60\sum_{i>j}(a_i\pm a_j)^8+6\sum_i(2a_i)^8
$$
Does anyone know if there`s a general formula for $N(a^2+b^2+c^2+d^2)^k$, for some $N$ depending on $k$, such that it is the sum of perfect powers of $2k$?
 [/INST] You should find the bounds of $a,b,c$ and show that $(a,b,c)=(1,2,3)$ and its permutations.
If you consider that $a\geq b\geq c$, then $(abc)^2 = a^3+b^3+c^3 \leq 3a^3\Leftrightarrow (bc)^2 \leq 3a$
Also, we see that $$a^2 \mid a^3+b^3+c^3\Rightarrow b^3+c^3 \geq a^2\Leftrightarrow bc^4 \leq 18$$
Therefore, $c=1$.
We now have to solve the equation $$a^3+b^3+1=a^2b^2$$
Of course $a\neq b$. If $b=2$, then $a=3$.
For $b\geq 2$, then $a>b+1$, so since $a^2|b^3+1$, we take that $a^2\leq b^2-b+1< b^2\Leftrightarrow a< b$, contradiction.
Therefore $$(a,b,c)=(1,2,3)$$ and its permutations.
</s>