The following values appear to work:
x=5, y=49, z=97
Alan: Please verify this. Thanks.
+23248 I agree with guest's answer of: x = 5, y = 49, and z = 97; this answer checks!
My analysis is this (but, I have no idea whether or not it is a correct procedure or only a lucky procedure):
1/x - 1/(xy) - 1/(xyz) = 19/97 with x < y < z
First find the sum of the left-side using the common denominator of xyz:
1(yz) / (xyz) - 1(z) / (xyz) - 1 / (xyz) = 19 / 97 ---> (yz - z - 1) / (xyz) = 19/97
The denominator of the left side has three factors; the denominator of the right side has only one factor, the prime number 97. It needs two more factors. Since z is the largest of the three factors, claim that = 97 and multiply the right hand side by xy/xy ---> (yz - z - 1) / (xyz) = 19xy / (97xy)
Claim that the numerators are equal: ---> yz - z - 1 = 19xy
and claim that the denominators are equal: ---> xyz = 97xy with z = 97.
Since z = 97 ---> yz - z - 1 = 19xy ---> 97y - 97 - 1 = 19xy
---> 97y - 98 = 19xy
---> 97y - 19xy = 98
---> y(97 - 19x) = 98
Since the factors of 98 are 2 and 49: ---> y(97 - 19x) = 2·49
So, if we are dealing with whole numbers,
either y = 2 or y = 49 (2 won't work ---> 49(97 - 19x) = 2·49
because y > x and x = 1 doesn't ---> 97 - 19x = 2
work). ---> -19x = -95
---> x = 5
