# OT Last week's CarTalk puzzler

Jim Hurley jhurley at infostations.com
Thu Nov 24 11:14:51 CST 2005

```>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

```