Recall that the ring of Gaussian integers is
$$\mathbb Z[i]=\{a+bi:\ a,b\in\mathbb Z\}.$$
Clearly
$$(a+bi)^2=a^2-b^2+2abi\ \ \mbox{and}\ \ (a+bi)^4=(a^2-b^2)^2-4a^2b^2+4ab(a^2-b^2)i.$$


**Question.** Is it true that $\{x^4+y^2+z^2:\ x,y,z\in\mathbb Z[i]\}=\{a+2bi:\ a,b\in\mathbb Z\}$?


**Evidence.** Via Mathematica I have found that
\begin{align} &\{x^4+y^2+z^2:\ x,y,z\in\{r+si:\ r,s\in\{-14,\ldots,14\}\}\}
\\&\quad \supseteq\{a+2bi:\ a,b\in\mathbb Z\ \mbox{and}\ |a+2bi|\le 50\}.
\end{align}
For example,
$$43+22i=2^4+(14-11i)^2+(11-13i)^2$$
and
$$-34+26i=(2+i)^4+(13-i)^2+(1+14i)^2=(4+i)^4+(11-11i)^2+(1+14i)^2.$$


**Motivation.** The question is motivated by my following result (cf. [my 2017 JNT paper](http://dx.doi.org/10.1016/j.jnt.2016.11.008))
$$\{x^4+y^2+z^2+w^2:\ x,y,z,w=0,1,2,\ldots\}=\{0,1,2,\ldots\}$$
which refines Lagrange's four-square theorem.


I conjecture that the question has a positive answer, but I'm unable to prove this. Your comments are welcome!



Is it true that $\{x^4+y^2+z^2:\ x,y,z\in\mathbb Z[i]\}=\{a+2bi:\ a,b\in\mathbb Z\}$?
Yes, it is true that $\{ x^{4} + y^{2} + z^{2} : x, y, z \in \mathbb{Z}[i] \} = \{ a + 2bi : a, b \in \mathbb{Z} \}$. Indeed, one can even take $x$ to be either $0$ or $1$ in all cases. Because $y^{2}+z^{2} = (y+iz)(y-iz)$ is reducible, this is analogous to the statement that every integer can be written in the form $x^{2}-y^{2}$ or $x^{2}-y^{2} + 1$ with $x, y \in \mathbb{Z}$.


Suppose that $a, b \in \mathbb{Z}$ and $a$ is odd. Then
$$
 a + 2bi = 0^{4} + \left(\frac{a+1}{2} + bi\right)^{2} + \left(b - \left(\frac{a-1}{2}\right)i\right)^{2}.
$$
If $a, b \in \mathbb{Z}$ and $a$ is even, then $a-1$ is odd, and the identity above allows one to write $a-1 + 2bi = y^{2} + z^{2}$ with $y, z \in \mathbb{Z}[i]$. Hence, $a + 2bi = x^{4} + y^{2} + z^{2}$ with $y, z \in \mathbb{Z}[i]$ and $x = 1$.


