For any positive integer n let f(n) denote the number of sets of four distinct positive integers a,b,c,d less than n such that ab-cd, ac-bd, and ad-bc are each multiples of n. In terms of congruences, f(n) is the number of sets of four distinct non-zero residues a,b,c,d such that ab=cd ac=bd ad=bc (mod n) Obviously f(n) is a measure of the divisibility of n, somewhat analogous to the sum proper divisors ("aliquot parts") of n, denoted by A(n). Recall that the ancient Greeks classified all the integers as either "deficient", "perfect", or "abundant" accordingly as A(n) is less than, equal to, or greater than n. Similarly we could define a different kind of perfection (and deficiency and abundance) according to whether f(n) is less than, equal to, or greater than n. According to this definition, the only "perfect" number I can find is 42, although proving that this is the only possible perfect number in this sense seems difficult. It would be helpful to have a simple expression for f(n) in terms of the factorization of n, but the problem is complicated by the fact that, unlike A(n)+n, the function f(n) is not multiplicative. (I've tried modifying the definition of f by allowing a,b,c,d to be 0 and/or non-distinct, but nothing seems to result in a multiplicative function.) The three basic congruences imply that, for any prime divisor p of n, if any one of the numbers a,b,c,d is divisible by p then at least three of them must be divisible by p. Also, multiplying the three congruences in various combinations gives (bc)a^2 = (bc)d^2 (bd)a^2 = (bd)c^2 (mod n) (cd)a^2 = (cd)b^2 so in the case where a,b,c,d are all coprime to p, it follows that a^2 = b^2 = c^2 = d^2 (mod p) Therefore, modulo any given prime divisor p of n, the set {a,b,c,d} must be of one of the three forms {0,0,0,0} or {k,0,0,0} (mod p) or {sqrt(k),sqrt(k),sqrt(k),sqrt(k)} Now, since we require a,b,c,d to be positive and distinct, it's clear that there are no acceptable quadruples if n is a prime or twice a prime (noting that a given residue k has at most two distinct square roots modulo a prime or twice a prime). Thus we have f(p) = f(2p) = 0 for every prime p. We can also see by inspection that f(9)=f(18)=0, but every other composite number n evidently has at least one solution. Here's a short list of the non-zero values of f(n) for all n less than 100: n f(n) n f(n) n f(n) n f(n) --- ---- --- ---- --- ---- --- ---- 8 1 36 20 57 9 80 291 12 2 39 6 60 238 81 97 15 2 40 65 63 36 84 356 16 3 42 42 64 133 85 16 20 4 44 10 65 12 87 14 21 3 45 24 66 70 88 161 24 33 48 147 68 16 90 288 25 1 49 15 69 11 91 18 27 9 50 26 70 84 92 22 28 6 51 8 72 311 93 15 30 28 52 12 75 161 95 18 32 31 54 102 76 18 96 1071 33 5 55 10 77 15 98 190 35 6 56 97 78 84 99 60 Taking the lazy approach of just looking for patterns, these numbers (and a few beyond this range) suggest the following rules for odd primes p, q, and r: f(p) = 0 f(2p) = 0 f(4p) = 1(p-1) + 0 f(8p) = 16(p-1) + 1 f(16p) = 72(p-1) + 3 f(32p) = 520(p-1) + 31 f(64p) = 2128(p-1) + 133 = 133 f(8p) f(128p) = 11904(p-1) + 981 f(pq) = (1/4)(p-1)(q-1) f(2pq) = (7/2)(p-1)(q-1) f(4pq) = (115/4)(p-1)(q-1) + f(4p) + f(4q) + f(pq) = 29(p-1)(q-1) + (p-1) + (q-1) f(9p) = 6(p-1) (except p=3) f(27p) = (429/2)(p-1) + 9 (except p=3) f(pqr) = (31/4)(p-1)(q-1)(r-1) + f(pq) + f(pr) + f(qr) Checking each of these forms for "perfection", we find that the only solution is for the case 2pq = f(2pq), which factors as (3p-7)(3q-7) = 28 with the solution p=3, q=7, giving the perfect number 42. It appears that the formulas for the remaining forms (e.g., f(27p) and f(pqrs) and so on) cannot give any more "perfect" numbers, but it would be nice to have a simple formula giving f(n) for ALL positive integers n. Does anyone know of such a formula? It seems to be some combination of the totients of the divisors of n, but I can't quite see it.

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