Find Triplets to Satisfy $x+y+z+xy+yz+zx=xyz+1$

[anemone]

Determine all triplets $(x,\;y,\:z)$ of positive integers such that $x \le y \le z$ and

$x+y+z+xy+yz+zx=xyz+1$.

 

[mente oscura]

Determine all triplets $(x,\;y,\:z)$ of positive integers such that $x \le y \le z$ and

$x+y+z+xy+yz+zx=xyz+1$.

Hello.

Spoiler

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

At a glance:

(1,0,0) \ (0,1,0) \ (0,0,1) \ (-1,-1,-1)

Only meets the two restrictions: (1,0,0)

Regards.

 

[eddybob123]

Spoiler

The above solution is not correct. The question asks for positive integers, and $0$ IS NOT POSITIVE!
$$2,4,13$$
$$2,5,8$$
$$3,3,7$$

 

[anemone]

Hello.

Spoiler

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

At a glance:

(1,0,0) \ (0,1,0) \ (0,0,1) \ (-1,-1,-1)

Only meets the two restrictions: (1,0,0)

Regards.

Thanks for participating, mente oscura! But your answer isn't correct. I'm sorry.:(

Spoiler

The above solution is not correct. The question asks for positive integers, and $0$ IS NOT POSITIVE!
$$2,4,13$$
$$2,5,8$$
$$3,3,7$$

Yes, those three are the only solutions but if you don't mind me asking, I would like to see how you approached the problem, sounds good to you?

 

[Opalg]

Determine all triplets $(x,\;y,\:z)$ of positive integers such that $x \le y \le z$ and

$x+y+z+xy+yz+zx=xyz+1$.

[sp]If $x$ (the smallest of these numbers) is $\geqslant4$ then each of $x,y,z$ is $\leqslant\frac14yz$; and each of $xy,xz,yz$ is $\leqslant yz$. Therefore $x+y+z+xy+yz+zx \leqslant \frac{15}4yz$. But $xyz + 1>4yz > \frac{15}4yz$. So there cannot be any soluions with $x\geqslant4.$

If $x=2$ then the equation becomes $2yz+1 = yz + 3(y+z) + 2$, so that $yz - 3(y+z) = 1$, or $(y-3)(z-3) = 10$. The only positive integer solutions with $y\leqslant z$ are $(y-3,z-3) = (1,10)$ or $(2,5)$, giving $(y,z) = (3,13)$ or $(5,8).$

If $x=3$ then the equation becomes $3yz+1 = yz + 4(y+z) + 3$, so that $yz - 2(y+z) = 1$, or $(y-2)(z-2) = 5$. The only solution is $(y-2,z-2) = (1,5)$, giving $(y,z) = (3,7)$.

Thus the only solutions are those given by eddybob:$(x,y,z) = (2,3,13),\ (2,5,8),\ (3,3,7)$.[/sp]

 

[anemone]

Thank you Opalg for participating and also your well explained solution!

Solution provided by other:

Spoiler

Let $x-1=p$, $y-1=q$ and $z-1=r$, the equation may be written in the form $pqr=2(p+q+r)+4$, where $p,q,r$ are integers such that $r \ge q \ge p \ge 0$. Observe that $p=0$ is not possible, for then $0=2(p+q)+4$ which is impossible in non-negative integers. Thus we may write this in the form

$2 \left( \dfrac{1}{pq}+\dfrac{1}{qr}+\dfrac{1}{rp} \right) +\dfrac{4}{pqr}=1$.

If $p \ge 3$, then $q \ge 3$ and $r \ge 3$. Then left side is bounded by $\dfrac{6}{9}+\dfrac{4}{27}$ which is less than 1.

We conclude that $p=1$ or $p=2$.

Case I:

Suppose $p=1$. Then we have $qr=2(q+r)+6$ or $(q-2)(r-2)=10$. This gives $q-2=1, r-2=10$ or $q-2=2$ and $r-2=4$ (recall $r \ge q$). This implies $(p,q,r)=(1,3,12), (1,4,7)$

Case II:
If $p=2$, the equation reduces to $2qr=2(2+q+r)+4$ or $qr=q+r+4$. This reduces to $(q-1)(r-1)=5$. Hence $q-1=1$ and $r-1=5$ is the only solution. This yields $(p,q,r)=(2,2,6)$

Reverting back to $x, y, z$ we get three triplets $(x,y,z)=(2,4,13),(2,5,8),(3,3,7)$.