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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 | Comment 1 of 7
Without having solved this, I might bet on the solution having something to do with the fact that the sum of the first N numbers, squared, equals the sum of the cubes of the first N numbers -- at least, that's what A and B remind me of.
 Posted by Old Original Oskar! on 2006-07-22 09:28:54

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