Q13/003) Find the unique triple (x,y,z) of positive integers such that x < y < z and (1/x)-(1/xy)-(1/xyz)=19/97

(1/x)-(1/xy)-(1/xyz)=19/97

=> 1- 1/y – 1/yz = 19x/97

Or 1/y + 1/yz = (97-19x)/97 so 19 x < 97 or x <=5

Now as 97 is in denominator on RHS so z has to be multiple of 97

If the value of z is chosen to be < 97 for simplicity and no solution with higher z can be proved

x = 1 => 1/y + 1/yz = 78/97 no solution as y has to be 2 and 1/2 + 1/6 = 2/3 < 78/97

x = 2 => 1/y + 1/yz = 59/97 no solution as 1/3 + 1/12  = 5/12 < 59/97

x= 3 => 1/y+ /yz = 40/97 no solution as 1/4 + 1/20 = 3/10 < 40/97

x= 4 => 1/y + 1/yz = 21/97 no solution as 1/5 + 1/(5* 97) = 98/(5 * 97) < 21/97

x =5 => 1/y + 1/yz = 2/97

or 97z + 97 = 2yz

or y = 97(z+1)/(2z)

if y = 97m then m = (z+1)/2z. m > 1 is not possible and m = 1 => z = 1 but z > y so no solution

z has to odd and multiple of 97

z= 97 => y = 49

z = 97 *3 => y = 146 and no solution for z > 97

so z = 97 and y = 49 is solution

so x = 5, y = 49 and z = 97 is the solution