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Does there exist a continuous map $f$ from $\mathbb C^4$ to $\mathbb R$ such that:

i) there exists four distinct complex numbers $a$, $b$, $c$, $d$, s.t. $f(a,b,c,d)f(b,c,d,a)<0$

ii) for every $(x,y,z,t)\in \mathbb C^4$, $f(x,y,z,t)=0$ implies that $x$, $y$, $z$, $t$ are the corners of a square in the complex plane?

This is related to the inscribed square problem.

Indeed, let's say that a polygon defined by $(a,b,c,d)$ is $f$-good if it satisfies the condition i), then an approach to solve the inscribed square problem can be to prove that there exists $f$ such that any Jordan curve contain a $f$-good polygon.

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  • $\begingroup$ How does $f$ being a polynomial ("polynome") justify the tag "model theory"? $\endgroup$
    – Wojowu
    Commented Apr 16, 2020 at 21:38
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    $\begingroup$ Whatever definissable sets are, I do not thing the tag is appropriate. $\endgroup$
    – Wojowu
    Commented Apr 16, 2020 at 21:52
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    $\begingroup$ @Gerry Myerson $f$ is real valued. $\endgroup$
    – jcdornano
    Commented Apr 17, 2020 at 0:46
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    $\begingroup$ @LSpice, thank you very much for the english grammar and presentation edit ! $\endgroup$
    – jcdornano
    Commented Apr 17, 2020 at 2:22
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    $\begingroup$ The idea is that if $abcd$ is not a square, one can find a path from it to $bcda$ such that no 4-gon on the path is a square. This is a contradiction because as require in the hypothesis $f(a,b,c,d)$ and $f(b,c,d,a)$ have a different sign, so $f$ continuous has to take the value $0$ for some 4-gone on the path, and this can happend only if the 4-gon is a square. $\endgroup$
    – jcdornano
    Commented Apr 17, 2020 at 11:05

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