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Cube first, or square second (Posted on 2006-07-22) Difficulty: 3 of 5
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)=81 and B=1+2+2+4= 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+...)=(a+b+c+...). Now, N.P^(Q-1) has (A, B, C...) as divisors, and also (AP, BP, CP...), (AP, BP, CP...), (AP, BP, CP,...) up to (AP^(Q-1), BP^(Q-1), CP^(Q-1)...).

If A has a divisors, AP has 2a divisors, AP 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+...)= Q(Q+1) times the original sum, so squaring it we would get a result Q(Q+1)/4 times greater as before.

Summing cubes, instead of (a+b+c+...) we now have (a+a.2+a.3+...a.Q+ b+b.2+b.3+...b.Q+ c+c.2+c.3+...c.Q+...) that is Q(Q+1)/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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