An Identity Concerning Averages
of Divisors of a Given Integer

Let me start with two definitions. Given a sequence of N > 0 numbers a1, a2, ..., aN, we define an arithmetic mean or average is defined

  A(a1, a2, ..., aN) = (a1 + a2 + ... + aN)/N

If none of the numbers is 0, we also define their harmonic mean by

  H(a1,a2,...,aN) = N/(1/a1 + 1/a2 + ... + 1/aN)

Let there be an integer M. We may consider all possible divisors of M, including 1 and M itself. We do not know how many there are. Nonetheless, for a given M, we may consider the arithmetic and harmonic means of the set of divisors. I denote them AM and HM, respectively. The task of computing either, especially the harmonic mean, may seem daunting. Thus it's all the more amusing to have the following

Theorem

M = AM · HM.

Examples

  1. M = 6 has four divisors: 1, 2, 3,and 6. A6 = (1 + 2 + 3 + 6)/4 = 3. On the other hand,

      H6 = 4/(1/1 + 1/2 + 1/3 + 1/6) = 4/(6/6 + 3/6 + 2/6 + 1/6) = 24/12 = 2.

    Therefore we indeed have 6 = 3·2.

  2. M = 8 has also four divisors: 1,2,4, and 8.

      A8 = (1 + 2 + 4 + 8)/4 = 15/4.
    H8 = 4/(1/1 + 1/2 + 1/4 + 1/8) = 4/(8/8 + 4/8 + 2/8 + 1/8) = 32/15.

    Finally, 15/4·32/15 = 32/4 = 8.

Proof

The main point here is to realize that the divisors of a number come in pairs: d divides M iff there exists an integer c such that M/d = c. This also can be written as

(1)1/d = c/M

Now, let's enumerate all the divisors of M as d1, d2,... We know that for every di there exists a ci such that di·ci = M for all i. Please note that ci is, by definition, also a divisor of M. Furthermore, as di ranges over the set of all divisors of M, so does ci. Let DM denote the number of divisors of M. From (1) we get

  HM = DM/(1/d1 + 1/d2 + ...) = DM/(c1/M + c2/M + ...) = M/AM.

Which exactly means that M = AM · HM.

References

  1. O.Ore Number Theory and Its History, Dover, 1988

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