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Given any $\beta \in R$, do there exist integers $k,l,m$, and a real number $x$ such that:

$l=4x^{3}m+6x^{2}m^{2}+4xm^{3}-2xm$,

as well as

$x^{4}-x^{2}=k+\beta$ ?

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    $\begingroup$ Yes, sure. Take $k=-\beta$ and $x=0$, with $l=m=0$. $\endgroup$ – Dietrich Burde Jun 13 '19 at 15:14
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    $\begingroup$ $\beta$ is not necessarily an integer, while $k$ is $\endgroup$ – Andrei Jun 13 '19 at 15:15
  • $\begingroup$ @DietrichBurde Doesn't work if $\beta\notin\mathbb Z$. $\endgroup$ – Adam Latosiński Jun 13 '19 at 15:16
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From the first equation, $x$ is algebraic. This implies $\beta$ algebraic, which might not hold.

There are countably many values of $\beta$ that give solutions. We can enumerate them by considering all triples $(k,l,m)$, finding the roots in $x$ and computing $\beta=x^4-x^2-k$.

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  • $\begingroup$ Yes, indeed. Thanks for pointing out that $\beta$ can only be countable. In general, sums and products of algebraic numbers are algebraic, and so we conclude the first statement. $\endgroup$ – Aritro Pathak Jun 13 '19 at 15:53

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