The 15x15 magic square is odd but also a multiple of 3. You can use the shift method to construct a 15x15 magic square but with boundary condition. Take as first row of the first and/or second grid 0-1-2-3-4-5-6-7- 8-9-10-11-12-13-14, than you get only a semi-magic 15x15 square. Take as first row of the first and/or second grid 0-2-1-3-4-5-8-7-6-11-10-9-13-12-14, and you get a panmagic 15x15 square.
 
The row 0-2-1-3-4-5-8-7-6-11-10-9-13-12-14 leads to a valid 15x15 panmagic square because 0+3+8+11+13 = 2+4+7+10+12 = 1+5+6+9+14 = 35, that is 1/3 of (0+1+2+3+4+5+6+7+8 +9+10+11+12+13+14=) 105.

 
 Take 1x digit from first grid (shift 2 to the left) +1

+15x digit from second grid (shift 2 to the right)

= panmagic 15x15 square