Solve this Diophantine equation: $4xyz=x+2y+4z$

Question

Solve this Diophantine equation: $4xyz=x+2y+4z$ $(x,y,z>0)$

My attempt: Without loss of generality, assume $x\ge y\ge z$

$=> 4xyz=x+2y+4z+3\ge7x+3$

At this I was stuck. I remember that I have solved $2xyz=x+y+z$ by that way and limited $y$ and $z$, but I can't limit this one.

First of all, there is loss of generality in that assumption, because your equation is not symmetric. — LHF, Feb 09 '20 at 07:44

score 1 · Accepted Answer · answered Feb 09 '20 at 07:55

We know that $x,y,z\geq 1$. The first step is to notice that $x$ is even. Now if $x \geq 3$, then:

$$4xyz=xyz+3xyz \geq x+9yz=x+2yz+7yz\geq x+2y+7z > x+2y+4z$$

That means $x=2$. And the equation becomes:

$$8yz=2+2y+4z\Rightarrow 4yz=1+y+2z$$

Now, similarly, if $y\geq 2$, we have:

$$4yz=yz+yz+2yz\geq 1+y+4z>1+y+2z$$

That means $y=1$. And solving for $z$, we get $z=1$. This gives the unique solution $(x,y,z)=(2,1,1)$.

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