xyz = 4(x + y + z) > 4z, so xy > 4.
xyz = 4(x + y + z) < 4(z + z + z) = 12z, so xy < 12
Combining these, 5 ≤ xy ≤ 11
If x = 1, then 5 ≤ y ≤ 11.
If x = 2, then 3 ≤ y ≤ 5. (as 5/2 and 11/2 are not integers.)
If x were 3, y would have to be less than 4, but y has to exceed x, so x can't be 3 or more.
If x = 1:
xyz = 4(x + y + z)
yz = 4(1 + y + z)
z(y - 4) = 4(y + 1)
z = 4(y + 1) / (y - 4)
Keeping within 5 ≤ y ≤ 11, y can be 5 (with z=24), 6 (with z=14) or 8 (with z=9); higher y would result in z lower than y.
If x = 2:
2yz = 4(2 + y + z)
yz = 4 + 2y + 2z
z(y - 2) = 2(y + 2)
z = 2(y+2) / (y-2)
Keeping within 3 ≤ y ≤ 5, y can be 3 or 4, with z being 10 or 6 respectively.
So the ordered triplets are (1,5,24), (1,6,14), (1,8,9), (2,3,10), (2,4,6). |
