Originally Posted by LittleCherub
( It doesn�t explain why all other non-perfect square numbers have even numbers of factors though.)
Erica already alluded to that: if f is a factor of n, that means n = kf for some integer k, so k = n/f is also a factor of n, and if f wasn't exactly the square root of n, then it's a different factor. So factors always appear in pairs, unless there is a square root. That's easier than using the prime factorisation, I think, although of course one can do it that way.

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