18
$\begingroup$

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) $$\{x^4+y^2+z^2+w^2:\ x,y,z,w=0,1,2,\ldots\}=\{0,1,2,\ldots\}$$ 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!

$\endgroup$
5
  • $\begingroup$ FWIW - some more evidence (with a short c++ program) if you let r,s above be in {-60, 60}, the only a, b you miss in [-130, 130]^2 is (6, -73) $\endgroup$ Nov 14, 2020 at 10:19
  • $\begingroup$ As you find a solution for $6+2\times73i$, there is a solution for $6-2\times73i$ (the conjugate of $6+2\times73i$) too. $\endgroup$ Nov 14, 2020 at 10:56
  • $\begingroup$ yes, I discarded symmetries $\endgroup$ Nov 14, 2020 at 11:03
  • $\begingroup$ Now I find a solution for $6-2\times73i$: $$6-146i=(2-i)^4+(61+6i)^2+(7-61i)^2.$$ $\endgroup$ Nov 14, 2020 at 11:30
  • $\begingroup$ As $(r-si)^4=(s+ri)^4$, we may restrict $x$ to the form $r+si$ with $r\ge0$ and $s\ge0$. To check the question efficiently, one may assume $x=r+si$ with $r,s$ relatively small. $\endgroup$ Nov 14, 2020 at 12:02

1 Answer 1

27
$\begingroup$

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 $x = 1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.