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Let $\mathbb Q_{\ge0}$ be the set of all nonnegative rational numbers. I have the following conjecture based on my computation.

4-3-2 Conjecture. Each $r\in\mathbb Q_{\ge0}$ can be written as $x^4+y^3+z^2$ with $x,y,z\in\mathbb Q_{\ge0}$.

As $m/n=(mn^{11})/n^{12}$ for any integers $m\ge0$ and $n>0$, it suffices to consider the 4-3-2 conjecture with $r\in\mathbb N=\{0,1,2,\ldots\}$. For example, \begin{align}2^{12}\times7=&2^4+15^3+159^2, \\4^{12}\times75=&122^4+1007^3+3951^2, \\3^{12}\times1140=&0^4+531^3+21357^2, \\5^{12}\times23710=&217^4+17897^3+232166^2. \end{align} For each $n=1,\ldots,30000$ with $n\not=23710$, I have found a number $m\in\{1,2,3,4\}$ such that $m^{12}n=x^4+y^3+z^2$ for some $x,y,z\in\mathbb N$.

QUESTIONS. Is the 4-3-2 conjecture true? Any way to prove it?

Your comments are welcome!

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  • $\begingroup$ Did you try numeric studies find any nonnegative integer with no 4-3-2 integer representation? You claim you needed $m=1,2,3,4$, can this set be reduced (on the bounded segment)? $\endgroup$
    – Ilya Bogdanov
    Commented Feb 3, 2022 at 22:05
  • $\begingroup$ If $m=1$ or $m=2$ is okay for $n$, then so is $m=4$. For $n=75$, $m$ cannot be $3$. For $n=1140$, $m$ cannot be $4$. For $n=23710$, $m$ must be greater than $4$. For more data, you may visit oeis.org/A350714 . For $n\le 40000$, we may require $m\le 4$ with the only exception $n=23710$ for which we may take $m=5$. $\endgroup$
    – Zhi-Wei Sun
    Commented Feb 3, 2022 at 23:30
  • $\begingroup$ @Ilya Bogdanov In 2015 I conjectured that $\{x^4-y^3+z^2:\ x,y,z=1,2,3,\ldots\}=\mathbb Z$. See oeis.org/A266152 . $\endgroup$
    – Zhi-Wei Sun
    Commented Feb 4, 2022 at 6:28
  • $\begingroup$ Another related conjecture of mine states that each integer $n>1$ can be written as $x^4+y^3+z^2+2^k$ with $x,y,z\in\mathbb N$ and $k\in\{1,2,3,\ldots\}$. See oeis.org/A280356 . $\endgroup$
    – Zhi-Wei Sun
    Commented Feb 4, 2022 at 6:34
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    $\begingroup$ Roth has proven that almost all positive integers are of the form $x^4+y^3+z^2$ with $x,y,z$ positive integers. $\endgroup$
    – Wojowu
    Commented Feb 6, 2022 at 14:14

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