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 Cube first, or square second (Posted on 2006-07-22)
For any number n, let

A= The square of the sum of the number of divisors of each of the divisors of n.

B= The sum of the cubes of the number of divisors of each of the divisors of n

Prove A=B

For example, let's have n=6. Its divisors are 1, 2, 3, and 6. These numbers respectively have 1, 2, 2, and 4 divisors. So A=(1+2+2+4)²=81 and B=1³+2³+2³+4³= 81.

 No Solution Yet Submitted by Jer Rating: 4.2857 (7 votes)

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 A bet winning solution | Comment 4 of 7 |
By induction: assume N (which isn't a multiple of prime P) satisfies the relationship, then prove that N.P^(Q-1) satisfies the relationship for all Q>0. The first step is easy, since N=1 does the job.

Now, N must have A, B, C... as divisors, and each of them has a, b, c... divisors itself. We have (a+b+c+...)²=(a³+b³+c³+...). Now, N.P^(Q-1) has (A, B, C...) as divisors, and also (AP, BP, CP...), (AP², BP², CP²...), (AP³, BP³, CP³,...) up to (AP^(Q-1), BP^(Q-1), CP^(Q-1)...).

If A has a divisors, AP has 2a divisors, AP² has 3a divisors, up to AP^(Q-1) that has Qa divisors, and the same happens to B, C... Thus, instead of (a+b+c+...) we now have (a+2a+3a+...+Qa+ b+2b+3b+...+Qb+ c+2c+3c+...Qc+...)= ½Q(Q+1) times the original sum, so squaring it we would get a result Q²(Q+1)²/4 times greater as before.

Summing cubes, instead of (a³+b³+c³+...) we now have (a+a.2³+a.3³+...a.Q³+ b+b.2³+b.3³+...b.Q³+ c+c.2³+c.3³+...c.Q³+...) that is Q²(Q+1)²/4 times the original sum of cubes.

Both results grow by the same amount, so the theorem is proved!

Edited on July 22, 2006, 1:33 pm
 Posted by Federico Kereki on 2006-07-22 13:27:55

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