## h0pe one year ago Find the unique triple $$x,y,z$$ of positive integers such that $$x < y < z$$ and $x=\frac{97yz-97z-97}{19yz}$.

1. anonymous

$x=\frac{ 97yz-97z-97 }{ 19yz }=\frac{ 97 }{ 19 }-\frac{ 97 }{ 19 yz }\left( y+1 \right)$ so $x<\frac{ 97 }{ 19 }$ but x is a positive integer. so x can be 1,2,3,4,5

2. anonymous

correction $\frac{ 97 }{ 19 yz }\left( z+1 \right)$

3. h0pe

what about y and z then

4. ganeshie8

$x=\frac{97yz-97z-97}{19yz} \implies 19xyz = 97(yz-z-1) \\~\\~\\ \implies z =\dfrac{97}{97y-19xy-97}$ Since $$z\gt 0$$ and $$z\gt y\gt x$$, it must be the case that $97y-19xy-97 = 1$ isolating $$y$$ : $y = \dfrac {98}{97-19x}$ so $$97-19x \in \{\text{positive factors of 98}\}$$

5. anonymous

yz-z-1>0 yz>z+1 yz>z yz-z>0 (y-1)z>0 y>1 so $y \ge2,z>y \ge2,z \ge3$

6. h0pe

I'm confused... how do I connect all of these equations

7. anonymous

Well, consider what ganeshie had. He had it worked down to y = 98/(97-19x), which means 97-19x had to be equal to a factor of 98. Factors of 98 are 2, 7, and 49. The only factor of 98 that makes the value of x equal to a positive integer is 2. 97-19x = 2 -19x = -95 x = 5 The other possible factors of 98 would fail. So since we can confirm x is 5, we would also get y to be 98/2 = 49 And from before $z = \frac{ 97 }{ 97y -19xy -1 }$ We confirmed that 97y -19xy -1 had to be equal to 1, which means z is 97.

8. h0pe

Should I put 97 into the equation now?

9. anonymous

if we try one by one no value of satisfy this . Hence no positive integers are possible with the given condition. Let us take the case x=3 57 yz=97 yz-97z-97 $40yz-97 z=97$ $z=\frac{ 97 }{ 40 y-97 }$ no integral value of y gives positive integral value of z

10. h0pe

the problem says there is an integral value of y that gives positive integral value of z

11. anonymous

in the statement it is also written x,y,z are positive integers

12. h0pe

If no integral value of y gives positive integral value of z then what o.o

13. anonymous

do you find anything wrong in my calculation?

14. h0pe

not really, no

15. h0pe

Well thanks, I guess