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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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 re(3): Solution Comment 7 of 7 |
(In reply to re(2): Solution by Bractals)

It looks correct to me now.  I do intend to do a further verification (what I would term an acid test) by specializing to n=(2^3)*(3^2) and working out each step numercially. I would urge others to try the same.
 Posted by Richard on 2006-07-23 04:00:33

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