Trimagic Square Impossibility Proofs The 3x3...6x6 impossibility proofs for normal bimagic squares also prove that normal trimagic squares of those sizes are impossible. Walter Trump has provided proofs for 9x9...11x11. Here are impossibility proofs for 7x7 and 8x8. 7x7 Normal Trimagic Square 8x8 Normal Trimagic Square 7x7 Normal Trimagic Square A 7x7 trimagic square of distinct integers in the range 1...49 is impossible. Proof The magic sum is 175, the magic sum of squares is 5775, and the magic sum of cubes is 214375. All three sums are 3 (mod 4) and 7 (mod 8). Since the magic sums are odd, a trimagic series must have an odd number of odd integers. But the sum of squares of 1 or 5 odd integers is not 3 (mod 4), so a trimagic series must have 3 or 7 odd integers. For a trimagic series with 3 odd integers, an odd number of the 4 even integers must be 2 (mod 4) to make the sum of squares 7 (mod 8). The even integers must be (0,0,0,2) (mod 4) or (0,2,2,2) (mod 4). The sum of the even integers is 2 (mod 4), so the 3 odd integers must sum to 1 (mod 4) to make the trimagic series sum to 3 (mod 4). The odd integers must be (1,1,3) (mod 4) or (3,3,3) (mod 4). Combining the even and odd integers makes four trimagic series. But in each, the sum of cubes of the even integers is 0 (mod 4) and the sum of cubes of the odd integers is 1 (mod 4) which do not sum to 3 (mod 4), as required. Thus, there is no trimagic series with 3 odd integers and 4 even integers, so a trimagic series must have 7 odd integers. But 7 rows would require 49 odd integers and there are only 25 in the range 1...49. 8x8 Normal Trimagic Square An 8x8 trimagic square of distinct integers in the range 1...64 is impossible. Proof The magic sum is 260 which is 2 (mod 3) and 8 (mod 9). The magic sum of squares is 11180 which is 2 (mod 3) and 2 (mod 9). The magic sum of cubes is 540800 which is 2 (mod 3) and 8 (mod 9). The square of 1 or 2 (mod 3) is 1 (mod 3). Let x stand for 1 or 2 (mod 3). For the sum of squares to be 2 (mod 3), the number of x's must be 2 more than a multiple of 3. (0,0,0,0,0,0,x,x) (mod 3) (x = 1,2) (0,0,0,x,x,x,x,x) (mod 3) (x = 1,2) (x,x,x,x,x,x,x,x) (mod 3) (x = 1,2) For the sum to be 2 (mod 3), the number of 2's among the x's must be a multiple of 3. This produces six trimagic series. (0,0,0,0,0,0,1,1) (mod 3) (0,0,0,1,1,1,1,1) (mod 3) (0,0,0,1,1,2,2,2) (mod 3) (1,1,1,1,1,1,1,1) (mod 3) (1,1,1,1,1,2,2,2) (mod 3) (1,1,2,2,2,2,2,2) (mod 3) The cube of 0 (mod 3) is 0 (mod 9), the cube of 1 (mod 3) is 1 (mod 9), and the cube of 2 (mod 3) is 8 (mod 9). This eliminates the following four of the six trimagic series because the sum of cubes is not 8 (mod 9), as required. (0,0,0,0,0,0,1,1) (mod 3) (0,0,0,1,1,1,1,1) (mod 3) (1,1,1,1,1,2,2,2) (mod 3) (1,1,2,2,2,2,2,2) (mod 3) In the trimagic series (1,1,1,1,1,1,1,1) (mod 3), each 1 (mod 3) is either 1, 4, or 7 (mod 9). Let a = number of integers of 1 (mod 9) in (1,1,1,1,1,1,1,1) (mod 3), b = number of integers of 4 (mod 9) in (1,1,1,1,1,1,1,1) (mod 3), and c = number of integers of 7 (mod 9) in (1,1,1,1,1,1,1,1) (mod 3). a + b + c = 8 integers in the trimagic series. a + 4b + 7c = 9j + 8 for some integer j, to get the magic sum. a + 7b + 4c = 9k + 2 for some integer k, to get the magic sum of squares. Combining the 1st and 2nd equations, 3b + 6c = 9j or 6b + 12c = 18j. Combining the 1st and 3rd equations, 6b + 3c = 9k - 6. Combining the two results, 9c = 18j - 9k + 6 or c = 2j - k + (2/3). Thus there is no solution in whole numbers. This leaves (0,0,0,1,1,2,2,2) (mod 3) as the only trimagic series. But 8 rows would require 24 integers of 0 (mod 3) and there are only 21 in the range 1...64.