A rectangular floor with integer sides (not necessarily distinct) has to be covered with...

For a rectangle with dimensions a × b, the number of edge (purple) tiles is 2a + 2b - 4, and interior (blue) tiles is (a-2)(b-2). Setting these equal gives (a-2)(b-2) = 2a + 2b - 4, which simplifies to ab = 4(a + b - 2). Testing each option: 8×1 (purple=14, blue=0, no); 5×3 (purple=12, blue=6, no); 4×4 (purple=12, blue=4, no); but for 5×5 (purple=16, blue=9, no). Only solutions are rectangles like 6×6 (purple=20, blue=16, no) or 5×5. Checking 16: if one side is 16, no valid integer second side satisfies the equation.

Step-by-step Derivation:
Let the rectangle have dimensions a × b (in units).

Total tiles = a × b
Purple tiles (perimeter) = 2a + 2b - 4 (corners counted once)
Blue tiles (interior) = (a-2)(b-2) = ab - 2a - 2b + 4

Condition: Blue tiles = Purple tiles
(a-2)(b-2) = 2a + 2b - 4
ab - 2a - 2b + 4 = 2a + 2b - 4
ab - 4a - 4b + 8 = 0
ab = 4a + 4b - 8
ab = 4(a + b - 2)

Rearranging: ab - 4a - 4b + 8 = 0
(a-4)(b-4) = 16 - 8 = 8

Valid factor pairs of 8: (1,8), (2,4), (4,2), (8,1)
This gives (a,b) pairs: (5,12), (6,8), (8,6), (12,5)

Verifying:

• 5×12: Purple = 2(5) + 2(12) - 4 = 34; Blue = (3)(10) = 30 ✗
• Let me recalculate: (a-4)(b-4) = 8 → valid (a,b): (5,12), (6,8), (8,6), (12,5)
• For 6×8: Purple = 20 + 16 - 4 = 32; Blue = (4)(6) = 24 ✗

Correct factorization: a=6, b=8 gives (6-4)(8-4)=2×4=8 ✓
Purple = 2(6)+2(8)-4 = 28; Blue = (4)(6) = 24 ✗

Rechecking from ab - 4a - 4b + 8 = 0:

• Option A (8): Possible with b=6 → 48 - 32 - 24 + 8 = 0 ✓
• Option B (5): Possible with b=12 → 60 - 20 - 48 + 8 = 0 ✓
• Option C (16): If a=16, then 16b - 64 - 4b + 8 = 0 → 12b = 56 → b = 4.67 (not integer) ✗
• Option D (12): Possible with b=5 → 60 - 48 - 20 + 8 = 0 ✓

Therefore, 16 units cannot be a side length.