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Let $z=(x+yi)$

$ x \in \mathbb{Z} , y \in \mathbb{Z} $

I am searching for complex numbers which satisfy following equations:

$|\Re(z^2)| \equiv |\Re(z)| \; ( \bmod \; b^{\lfloor\log_{b}(|\Re(z)|)+1\rfloor} ) \wedge $

$ |\Im(z^2)| \equiv |\Im(z)| \; ( \bmod \; b^{\lfloor\log_{b}(|\Im(z)|)+1\rfloor} ) \; , b \in \mathbb{N} $

Which is:

$ |x^2 - y^2| \equiv |x| \; ( \bmod \; b^{\lfloor\log_{b}(|x|)+1\rfloor} ) \wedge$

$ |2xy| \equiv |y| \; ( \bmod \; b^{\lfloor\log_{b}(|y|)+1\rfloor} )$

The basic idea about these modulo equations is that $x^2−y^2$ ends with the digits of $x$ and $2xy$ ends with the digits of $y$, ignoring the signs. e.g. $−123$ ends with $23$.

$b$ is a base. I would be happy if it could be solved with $b=10$. But it would be even better if it could be solved for any base $b$.

An example of a complex number in base $b=10$:

(313 + 216i)² = 51313 + 135216i

So far, I have found 1447 numbers using a brute-force algorithm, but it is stuck and I wonder if there is more to find or if the amount of numbers was finite.

The smallest found number is $(8 + 4i)$, the biggest found number is $(959106445313 + 749939593216i)$. For some reason, extremely many numbers have $\Im(z)=60406784$.

Is there a rule or algorithm to find all possible $(x,y)$ tuples? Are there infinite numbers which satisfy these equations?

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  • $\begingroup$ ??? What is $b$? $\endgroup$ – Robert Israel Jul 15 at 15:12
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    $\begingroup$ What is the absolute value in the modulo world? $\endgroup$ – Christian Blatter Jul 15 at 15:16
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    $\begingroup$ @ChristianBlatter My guess is that by $|s|\equiv |t| \mod m$ the OP could mean: $s\equiv t \mod m$ or $s \equiv -t \mod m$. $\endgroup$ – Robert Israel Jul 15 at 15:48
  • $\begingroup$ Sorry, I forgot to define $b$. It is a base. I would be happy if it could be solved with $b=10$. But it would be even better if it could be solved for any base $b$. The basic idea about these modulo equations is that $2xy$ ends with the digits of $y$, and $x^2 - y^2$ ends with the digits if $x$, ignoring the signs. e.g. $-123$ ends with $23$. $\endgroup$ – Daniel Marschall Jul 15 at 17:15

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