Talk:Powerful number

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This article includes too many statements without derivation, "clearly"s, and "since a, b" which assumes the reader knows the property being used.--Walt 19:26, 22 June 2006 (UTC)

Untagging since I believe the edits I made today address this complaint. —David Eppstein 00:31, 12 September 2006 (UTC)

[edit] Confusion about the opening definition

A powerful number is a positive integer m that for every prime number p dividing m, p2 also divides m. Equivalently, a powerful number is the product of a square and a cube, that is, a number m of the form m = a2b3, where a and b are positive integers.

How does the second sentence follow from the first? Surely the definition in the first sentence is satisfied by any square number? Colonies Chris (talk) 13:38, 25 February 2008 (UTC)

Any square number also satisfies the second definition, with b = 1. In general, one can take b to be the product of the prime numbers that have an odd exponent in the prime factorization of m, and a = √m/b3. —David Eppstein (talk) 15:55, 25 February 2008 (UTC)
But the first sentence completely rules out odd-exponent primes. It says, if p divides m, so does p2. That means that all the prime factors of m have to have even exponents. (Of course some of the exponents might be divisible by three as well, but nothing in that sentence requires it or even hints at it). Now I'm even more baffled. Colonies Chris (talk) 12:14, 27 February 2008 (UTC)
p2 divides p3. Why would you think it doesn't? —David Eppstein (talk) 15:21, 27 February 2008 (UTC)
Sorry, I had got it into my head that the exponent of each prime factor had to be even, rather than simply >= 2. Now it makes sense. Could the explanation of the equivalence be done less 'technically' by considering each prime factor's exponent modulo 6? (0 & 3 are cubes, 2 and 4 are squares, 5 is a square times a cube, 1 is a cube times a square provided the exponent is > 1, which it always is). Colonies Chris (talk) 12:19, 28 February 2008 (UTC)