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$x$, $y$ and $z$ are prime numbers. Solve the following equation: $$105 (x + y + z) = xyz$$

Knowing that $$105 = 3 × 5 × 7$$ it seems that there's no solution for this equation considering that $xyz$ has to be divisible by 3, 5 and 7 and the only prime numbers divisible by 3, 5 and 7 are 3, 5 and 7.

Thanks in advance!

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    $\begingroup$ Do you consider numbers like $-7$ primes? $\endgroup$
    – Hagen von Eitzen
    Nov 29, 2020 at 22:34
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    $\begingroup$ Related? math.stackexchange.com/questions/3920034/… $\endgroup$
    – Hagen von Eitzen
    Nov 29, 2020 at 22:48
  • $\begingroup$ I don't. Only positive numbers greater than 1. $\endgroup$
    – Moz
    Nov 29, 2020 at 22:49
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    $\begingroup$ Your equation has many integer solutions, for example $$(x,y,z)=(-18,-15,-21),(9,-40,7),(18,15,21)$$ but your reasoning about the impossibility of prime solutions is correct. $\endgroup$
    – Piquito
    Nov 29, 2020 at 22:53

1 Answer 1

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For $x\in\mathbb{Z}$, let $\Omega(x)$ be the number of divisors of $x$ that can be written as $p^i$ where $p$ is a prime number and $i\geqslant 1$. Then if there is a solution for your equation, we have $\Omega(xyz)=\Omega(105(x+y+z))$, that is to say $3=3+\Omega(x+y+z)$ and thus $\Omega(x+y+z)=0$ which leads to the contradiction $x+y+z=1$. You can also, with your argument $105|xyz$, say that $\{x,y,z\}=\{3,5,7\}$ and thus $x+y+z=1$.

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