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
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Posted by broll
on 2023-09-18 22:36:57 |
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