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Consider the diophantine equation $$ x^n+y^n+z^n=k\cdot xyz, $$ where $n,x,y,z$ are positive integers. Now consider $k\in\left\{4,7,8\right\}$. It is well-known or easily provable that for $n=1$ and $n=2$ the equation is unsolvable. This fact remains true for $n=3$, this time the proof is more elaborated and can be found in the paper Third-degree diophantine equations by M. Z. Garaev, see the link http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.36.8786 Hence, arise the natural question:

Q. This equation is solvable for $n\geq 4$ and $k=4$? The same question for $k=7$ and $k=8$.

Thanks in advance.

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    $\begingroup$ If $x,y,z$ are positive then $x^n + y^n + z^n \geq 3 (xyz)^{n/3}$ (AM-GM inequality), so once $n>3$ it's a finite search. $\endgroup$ Apr 4, 2015 at 15:58
  • $\begingroup$ Thanks @Noam D. Elkies, a priori $n$ could be large. We have $xyz\leq \left(\dfrac{k^3}{27}\right)^{\frac{1}{n-3}}$. For $k=4$ and $n=4$ it follows $xyz=1$ or $xyz=2$. In this case the equation is unsolvable. $\endgroup$
    – user70097
    Apr 4, 2015 at 16:12
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    $\begingroup$ But as $n$ increases, your bound on $xyz$ becomes more stringent. So if you're really only interested in $k\le 8$, then as soon as $n\ge8$, you get $xyz<2$ so $xyz=1$. So you're reduced to checking for solutions with $n\le 7$, which as Noam points out, is a finite search. Indeed, for $k=8$ you just need to check $xyz\le2$ for $n=6$ and 7, $xyz\le4$ for $n=5$, and $xyz\le18$ for $n=4$. Only the latter will take any time, and it is quite feasible on a computer. $\endgroup$ Apr 4, 2015 at 16:24
  • $\begingroup$ Thanks @JoeSilverman, the question is answered. Supposing $x\leq y\leq z$, we have $x^3\leq xyz\leq 18$, hence $x=1$ or $x=2$. $\endgroup$
    – user70097
    Apr 4, 2015 at 16:50
  • $\begingroup$ Now a more interesting question is what occur with values $k\in \left\{4l,8l-1,2^{2t+1}(2l-1)+3\right\}$. According to Garaev's paper the equation is unsolvable for $n=3$. For $n=1$ the equation is only solvable for $k\in\left\{1,2,3\right\}$ and for $n=2$ only for $k\in\left\{1,3\right\}$. $\endgroup$
    – user70097
    Apr 4, 2015 at 16:58

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