Charles Hartman wrote:
Interesting. At first it looks straightforward:
1. If a factor is by definition an integer that when multipilied by
another integer
yields the number we're interested in as a product, then factors
have to come
in pairs. (It takes two to multiply.)
2. "Odd number of factors" is therefore a contradiction in terms,
unless "factor" is
shifted to mean "unique factor".
3. If a number has a pair of factors that are identical (so not
unique, so they only
"count" once), then it's the product of that factor (which provably
can't be
either 1 or the number itself), times that factor, which is the
definition of a
square.
So "having an odd number of factors" is a sufficient condition for
being "a square".
But it doesn't seem to be a necessary condition. The factors of 36 --
by the double definition you have to use in order to make sense of
the statement of the problem -- are either
1 36 2 2 3 3
or
1 36 2 3
Charles,
The factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, and 36; an odd number
of factors as the theorem predicts.
The factors of 32 are 1, 2, 4, 8, 16, and 32; an even number of
factors, again consistent with the theorem.
The only way I know to solve this problem, i.e. show that the number
of factors of a given number is odd, if and only if the number is a
perfect square, is to express the number as a product of primes.
For example 36 = 2^2 * 3^2
And 32 = 2^5
In case anyone wants to pursue this further, I will quit here. Beauty
is best enjoyed if you discover it for yourself.
Jim
_______________________________________________
use-revolution mailing list
use-revolution@lists.runrev.com
Please visit this url to subscribe, unsubscribe and manage your subscription
preferences:
http://lists.runrev.com/mailman/listinfo/use-revolution
