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.
(In reply to
A bet by Old Original Oskar!)
You would win that bet. I used that and the Unique Factorization Theorem to solve the problem. The only thing is that the proof is loaded with superscripts and subscripts. Here would be a nice place for the comments section to allow HTML tags.
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Posted by Bractals
on 2006-07-22 09:39:17 |