The set of integers {x,y,z} complies with the following set of equations:
x+y+xy=17
x+z+xz=35
y+z+yz=71
Evaluate {x,y,z}.
Generalising over the integers:
Using the substitution (a+b+ab+1)= (a+1)(b+1)
let x,y,z be integers such that x+y+xy+1=k, x+z+xz+1=2k, y+z+yz+1=4k
2(x+z+xz+1)=(y+z+yz+1), then (z+1) (2x y+1)=0, with a solution at y=(2x+1) when k=2(x+1)^2
2(x+y+xy+1)= (x+z+xz+1), then (x+1) (2y z+1)=0, with a solution at z=(2y+1) when 2k=4 (x+1)^2
4(x+y+xy+1)= (y+z+yz+1), then (y+1) (4 x z+3)=0, with a solution at z=(4x+3) when 4k=2(y+1)^2 = 8(x+1)^2
When k=18, x={4,2}, y={7,5},z={13,11}
Lastly, if (y=1), or (z=1) then x would be less than 1, so the other possibility is (x=1), but if k?=0, (x=1), then 1+y+1y+1=0, but this is a contradiction.
So the required solutions are {x,y,z}={4,7,13}, {2,5,11}
Edited on September 18, 2023, 10:47 pm

Posted by broll
on 20230918 22:36:57 