I've done some work on (semi)magic squares whose elements' reciprocals also form (semi)magic squares. I'm calling them Sallows squares, for Lee Sallows, who introduced the idea here on October 12th. Lemma 1: If a+b = c+d and 1/a + 1/b = 1/c + 1/d, a, b, c, and d being different from each other, then two of the four numbers must be the negatives of the other two. Proof: There must exist a non-zero n such that c = a+n and d = b-n. Then 1/a + 1/b = 1/(a+n) + 1/(b-n) = ((a+n)+(b-n)) / (a+n)(b-n) = (a+b)/(a+n)(b-n) = (a+b)/(ab+bn-an-nn) = (a+b)/(ab + n(b-a-n)). 1/a + 1/b = (a+b)/(ab), so (a+b)/(ab) = (a+b)/(ab + n(b-a-n)). If a+b is non-zero, we can take the reciprocal of both sides, then multiply by a+b, to get ab = ab + n(b-a-n), so n(b-a-n) = 0. Since n != 0, b-a-n = 0. But since b-n = d, a must equal d, contradicting the claim that a, b, c, and d were distinct. So a+b must be 0. (And, by symmetry, c+d must also be 0.) QED. Lemma 2: If a+b+c+d = 0 and 1/a + 1/b + 1/c + 1/d = 0, with a, b, c, and d being real and being different from each other, then two of the four numbers must be the negatives of the other two. Proof: Since the sum is zero, the numbers can't all have the same sign. So either one is is one sign and the other three are the other, or it's two and two. Two and two with two of the four numbers not being the negatives of the other two is ruled out by lemma 1, since if, for example a and b are positive and c and d are negative, a+b+c+d = 0 would mean a+b = c+d, and 1/a + 1/b + 1/c + 1/d = 0 would mean 1/a + 1/b = 1/c + 1/d. So that leaves one number with one sign and the other three with the other. Choose any two of the four numbers with opposite signs. If their absolute values are equal, then they're negatives of each other, and since the other two numbers have the same sign as each other, they can't add up to 0. (They can't *be* 0, since 0 has no reciprocal. And also because they would be duplicates.) So the two you chose have different absolute values. Whichever number has the larger absolute value, the remaining two numbers would have to have the opposite sign to make a+b+c+d = 0 work, but would have to have the same sign to make 1/a + 1/b + 1/c + 1/d = 0 work. For instance if the numbers you chose were 1 and -6, to make the latter equation work the other two numbers might be -3 and -2, but to make the former equation work, the other two numbers might be +3 and +2. So one with one sign and three with the other can't work either. QED. Theorem 1: An order-3 semimagic (or magic) Sallows square is impossible. Proof: Consider the semimagic square: ABC DEF GHI A+B+C = A+D+G, and 1/A + 1/B + 1/C = 1/A + 1/D + 1/G, so B+C = D+G, and 1/B + 1/C = 1/D + 1/G, so, by lemma 1, B+C = 0. But, by left- right symmetry, A+B also equals 0. Since A+B = B+C then A=C, violating uniqueness. QED. Theorem 2: A real-number order-4 magic Sallows square is impossible. Proof: Consider the magic square ABCD EFGH IJKL MNOP with sum S. Add the top and bottom rows and the main diagonal, and subtract from that the middle two columns and the other diagonal: (A+B+C+D) + (M+N+O+P) + (A+F+K+P) - ((B+F+J+N) + (C+G+K+O) + (D+G+J+M)) = 0. After canceling, you get (2A + 2P) - (2G + 2J) = 0, i.e. A+P = G+J. But it must also be the case that 1/A + 1/P = 1/G + 1/J, so by lemma 1, A = -P and G = -J. By symmetry, D+M = F+K and 1/D + 1/M = 1/F + 1/K, so D = -M and F = -K. That means the diagonal sums are both 0, so the magic sum is 0. It's easy to construct such a magic square, for instance 1 -5 -3 7 -2 6 4 -8 8 -4 -6 2 -7 3 5 -1 but the reciprocals can't be made to work, except on the diagonals. Consider the top row, A,B,C,D. None of the numbers on that row can be the negative of A, since that value is taken by P. None of the numbers on that row can be the negative of D, since that value is taken by M. So A and D must have different absolute values. As such, B and C can't be negatives of each other, since if they were they'd sum to 0, in which case A+B+C+D wouldn't sum to 0. So lemma 2 shows that there's no solution to that row. QED. (The same argument applies to every row and every column, but it only takes one bad row or column to ruin the square.) I haven't proven that an order-4 magic square in complex numbers is impossible, though I believe it is. Perhaps one of you can extend my proof to cover complex numbers. Note that lemma 1 applies to complex numbers, but lemma 2 does not. I haven't proven that a real order-4 semimagic square is impossible. I've been trying to find one in positive rationals by an extensive search. First, note that if there's a solution in rationals, there's a solution in integers, since the rationals can all be multiplied by their common denominator without disturbing either the magic square or the reciprocal magic square (though it may change their magic sums). My method is, for each positive integer sum, find all possible sets of four numbers that add up to that sum, then find all sets of at least eight such sets whose reciprocal sums are equal to each other. Then I try to assemble eight of them into a semimagic square. Here's the best I've found so far: {100,110,450,1100}, {104,120,288,1248}, {60,408,612,680}, {60,420,560,720}, {60,450,500,750}, {66,231,693,770}, {66,264,440,990}, {70,195,585,910}, {70,234,364,1092}, {72,176,720,792}, {79,180,316,1185}, {80,144,576,960}, {80,180,300,1200}, {85,136,459,1080}, {86,126,645,903}, {90,170,225,1275}, {95,126,342,1197}, {99,121,330,1210}. Those 18 sets of four numbers all sum to 1760, and their reciprocals all sum to 1/45. Good luck finding better; I looked at more than 40 billion sets of four numbers to find those. But it's no good. As with every such set of sets that I've found, each set of four contains at least one number not found in any of the other sets. For instance in {100,110,450,1100} above, only 450 appears in another of the 18 sets of sets. Obviously for any permutation of {100,110,450,1100} to be a row or column of a semimagic square, all four of the numbers must appear in other sets. (At least lemma 1 guarantees that if they do, they'll be in four different sets, which is what we want.) Maybe there's a proof that every such set must contain at least one number not found in any other set. If so, a semimagic order-4 Sallows square in positive rationals is impossible. Or maybe it's just a matter of probabilities, given that those numbers seem to be almost randomly chosen from the numbers less than the target sum S. (There does seem to be a bias toward numbers with lots of divisors (e.g. 47 of the 63 distinct numbers in that set of sets are even). Also, I'm excluding all numbers divisible by a prime greater than S/4, as those can't possibly be part of a Sallows square. Can you see why not?) If it is just a matter of probabilities, then finding such a semimagic square without insights to greatly narrow the search would be almost intractable. I haven't yet searched for a semimagic square in mixed signs, only in positives. I haven't yet done anything with orders greater than four (except for the order-six complex-number semimagic square I posted on October 15th).