I guess I can assume you have all had time to work on this beautiful
problem. But if not, read no further.
I really like the solution that Charles Hartman provided. Simple and precise.
It wasn't the solution I had in mind at all. There is another
approach, not as quick as Charles', but it not only demonstrates
when the number of factors is odd or even, but it also provides a
method of obtaining how many odd or how many even factors there are,
and, as a bonus, as a process for finding those factors.
From the fundamental theorem of arithmetic any number N can be
represented as a unique product of primes
N = p1^n1 * p2^n2 * .... etc.
where p1, p2, etc are prime numbers, and n1, n2, etc. are integers
The factors of N are then
p1^m1 * p2^m2 * ..... etc.
where m1 takes on values from 0 to n1, m2 takes on values from 0 to n2, etc.
The number of factors is therefore: (n1 + 1) * (n2 + 1) etc.
This product will be odd if and only if EACH of the brackets is odd;
But (n + 1) is odd if and only if n is even.
If all the n's are even, it follow that N is as perfect square.
Similarly for the odd number of factors.
For example, if N = 24 then its prime factor representation is
24 = 2^3 * 3^1
And the number of factors is the exponent 3 plus 1, times the
exponent 1 plus 1, or 4 * 2 = 8
Explicitly the 8 factors are:
1,2,3,4,6,8,12,24
Each factor is 2 to some power less than or equal to 3, multiplied by
3 to some power less than or equal to 1.
Jim
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