- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,680

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

- Thread starter anemone
- Start date

- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,680

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

- Nov 29, 2013

- 172

Hello.

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

At a glance:

[tex](1,0,0) \ (0,1,0) \ (0,0,1) \ (-1,-1,-1)[/tex]

Only meets the two restrictions: [tex](1,0,0)[/tex]

Regards.

- Aug 18, 2013

- 76

$$2,4,13$$

$$2,5,8$$

$$3,3,7$$

- Thread starter
- Admin
- #4

- Feb 14, 2012

- 3,680

Thanks for participating,Hello.

At a glance:

[tex](1,0,0) \ (0,1,0) \ (0,0,1) \ (-1,-1,-1)[/tex]

Only meets the two restrictions: [tex](1,0,0)[/tex]

Regards.

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?

$$2,4,13$$

$$2,5,8$$

$$3,3,7$$

- Moderator
- #5

- Feb 7, 2012

- 2,703

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

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

- Thread starter
- Admin
- #6

- Feb 14, 2012

- 3,680

Solution provided by other:

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

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