What are all the ordered triples of positive integers (x,y,z) with x<y<z, such that their product is four times their sum?

For x = 1;

yz - 4y - 4z = 4 or (y - 4)(z - 4) = 20, giving (1,5,24), (1,6,14), (1,8,9) as solutions.

For x = 2;

yz - 2y - 2z = 4 or (y - 2)(z - 2) = 8, giving (2,3,10), (2,4,6) as solutions.

Now x = (4y + 4z)/(yz - 4).

For x>=3

3yz - 12 <= 4y + 4z

9yz - 12y - 12z <=36

(3y - 4)(3z - 4) <= 52

(3y - 4)^2 <= 52

3y - 4 <= 7

y <= 3

This last contradicts the conditions of the problem, so all solutions have been found.

Posted by xdog on 2007-11-24 13:22:23