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

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The discussion centers around finding all triplets of positive integers (x, y, z) that satisfy the equation x+y+z+xy+yz+zx=xyz+1, with the condition x ≤ y ≤ z. It is established that if x is greater than or equal to 4, no solutions exist due to the inequality derived from the equation. For x equal to 2, the solutions (y, z) are found to be (3, 13) and (5, 8), while for x equal to 3, the only solution is (3, 7). The confirmed solutions are (2, 3, 13), (2, 5, 8), and (3, 3, 7). The discussion concludes with acknowledgment of contributions from participants.
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Determine all triplets $(x,\;y,\:z)$ of positive integers such that $x \le y \le z$ and

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

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

(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.:(

eddybob123 said:
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?
 
anemone said:
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]
 
Thank you Opalg for participating and also your well explained solution!

Solution provided by other:

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 $(x,y,z)=(2,4,13),(2,5,8),(3,3,7)$.
 
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