Problem Statement

For this task, for our purposes, a cuboid is a 3 dimensional object, with six rectangular faces, where all angles are right angles, opposite faces of the cuboid are equal, and where each dimension is a positive integer unit length. It will subsequently be referred to simply as a cuboid; but be aware that it references the above definition. The surface area of a cuboid is two times the length times the width, plus two times the length times the height, plus two times the width times the height. A cuboid will always have an even integer surface area. The minimum surface area a cuboid may have is 6; one where the l, w, and h measurements are all 1. Different cuboid configurations (may) yield different surface areas, but the surface area is always an integer and is always even. A cuboid with l = 2, w = 1 h = 1 has a surface area of 10 There is no configuration which will yield a surface area of 8. There are 16 known even integer values below 1000 which can not be a surface area for any integer cuboid. It is conjectured, though not rigorously proved, that no others exist. that can not be the surface area of a cuboid.

Let's start with the solution:

int  L, W, H, HA, I;
char T(1000/2);                 \table of half areas
[for I:= 0 to 1000/2-1 do
        T(I):= true;            \assume all are O'Halloran numbers
for L:= 1 to 250 do
    for W:= 1 to 250/L do
        for H:= 1 to 250/L do
            [HA:= L*W + L*H + W*H;
            if HA < 500 then    \not an O'Halloran number
                T(HA):= false;
            ];
for I:= 6/2 to 1000/2-1 do
    if T(I) then
        [IntOut(0, I*2);  ChOut(0, ^ )];
]