Is Every Number with an Odd Number of Divisors a Perfect Square?

[GeoMike]

Ok, the proof to be done is pretty simple:
Prove that a number is a square only when the number of positive divisors is odd.

I pretty sure I know the answer, I'm just not sure how to go about writing it out...

If c is the number then you can write:
c=ab.
a and b are divisors of c. If a doesn't equal b then you have two different divisors. If a=b, then c=ab can be rewritten as c=a^2, and you only have one divisor. Because of this any number that can be written as a square of another number has an odd number of divisors -- all the pairs of factors that equal the number plus the 1 divisor that is squared to make the number.

I guess I just need to know how to write this out better, and how to make sure I haven't assumed to much as given/proven at the outset.
Thanks,
-GM-

 

[shmoe]

That's pretty much it, you've got the essential idea of pairing divisors c=ab with a,b distinct, which leaves out the oddball $$\sqrt{c}$$ out. You could try organizing the divisors c=ab with the assumption that a<=b, but this isn't necessarily any better.

For an alternate way, you could look at the prime factorization of c and count the divisors that way. (this isn't a 'better' way, but gives a different point of view to consider if you run across other divisor problems)