Differently Perfect
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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