Find all triples of natural numbers $x$, $y$, $z$ such that $x+2y+2z=xyz$.

Question

Find all triples of natural numbers $x$, $y$, $z$ such that $x+2y+2z=xyz$.

I found a solution, but it involves a lot of case work. Can someone help me find a solution which doesn't involve a lot of case work?

Answer 1

We have: $y = \dfrac{x+2z}{xz-2}\ge 1\implies x+2z\ge xz-2\implies 2z+2\ge xz-x = x(z-1)\implies x \le \dfrac{2z+2}{z-1}$. Of course the case $z = 1$ warrants a separate consideration and it's not hard to solve for this case as you have down to $2$-variable equation in $x,y$ only and you can isolate either $x$ or $y$ and take care of the case when a fraction is an integer. So for $z > 1$, then $x \le \dfrac{2z+2}{z-1} \le 3$ when $z \ge 5$. Thus you have to consider cases $z = 1,2,3,4$ first as well. And when $z \ge 5$, you have $x = 1,2,3$ and treat each case again. So in brief, it seems that's what amounts to and you have to be willing to work hard this time to treat about $10$ cases !

Answer 2

Let $x$, $y$ and $z$ be natural numbers satisfying $$x+2y+2z=xyz.$$ If $xyz=0$ then $x=y=z=0$, so suppose $x,y,z>0$. A bit of algebra shows that $$(xy-2)(xz-2)=x^2+4.\tag{1}$$ Of course without loss of generality $y\leq z$. If $x=1$ then the above simplifies to $$(y-2)(z-2)=5,$$ and so $y=3$ and $z=7$. Otherwise $x\geq2$ and then from $y\leq z$ it follows that $$(x+1)^2>x^2+4=(xy-2)(xz-2)\geq(xy-2)^2,$$ and hence that $x\geq xy-2$, or equivalently $2\geq x(y-1)$. This means that either $x=y=2$, in which case $z=3$, or $y=1$, in which case $(1)$ simplifies to $$(x-2)(xz-2)=x^2+4.\tag{2}$$ Clearly $x\neq2$, and so from the identity $$xz-2=\frac{x^2+4}{x-2}=x+2+\frac{8}{x-2},$$ we see that $x-2$ divides $8$, so $x-2\in\{1,2,4,8\}$. Plugging these values into $(2)$ yields \begin{eqnarray*} 1\cdot(3z-2)&=&13,\\ 2\cdot(4z-2)&=&20,\\ 4\cdot(6z-2)&=&40,\\ 8\cdot(10z-2)&=&104. \end{eqnarray*} The first three have the solutions $z=5$, $z=3$ and $z=2$, respectively, and the last has no integral solution.

In summary, up to permutation of $y$ and $z$, the solutions $(x,y,z)$ are $$(0,0,0),\quad(1,3,7),\quad(2,2,3),\quad(3,1,5),\quad(4,1,3),\quad(6,1,2).$$

Answer 3

"OP" equation is:

$x+2y+2z=xyz$ ---(1)

Eqn. (1) has parameteric solution's shown below.

$(x,y,z)=(3k^2+1)/(k),(3k),(1/k)$

For, $k=1$, we get: $(x,y,z)=(4,3,1)$

& for, $k=1/3$, we get: $(x,y,z)=(4,1,3)$

Another parametrization is:

$(x,y,z)=(4k^2+2)/(k),(2k),(1/k)$

For, $k=1$, we get: $(x,y,z)=(6,2,1)$

& for, $k=1/2$, we get: $(x,y,z)=(6,1,2)$