# Solve the following equation if x, y and z are prime numbers

$$x$$, $$y$$ 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.

• Do you consider numbers like $-7$ primes? Nov 29, 2020 at 22:34
• 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. Nov 29, 2020 at 22:53
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 $$x+y+z=1$$.