Determine all positive whole numbers w such that d(w) = √w, where d(w) denotes the number of positive divisors of w.
Let w = π(pi)^2*ki in prime factorization form.
(As d(w) is a whole number,so is √w and w is a perfect square)
So, d(w) is odd because d(w)=π(2*ki+1). So w is odd.
Given √w/d(w)=1 => π((pi)^ki)/(2*ki+1)=1
Take f(k)=(p^k)/(2*k+1), f(0)=1 for every p, its a monotonically
increasing function for p>=5 (which is prime) and p^k/(2*k+1)>1
only for p=3,f(1)=3/(2*1+1) =1, for k>1, P^k/(2*k+1) >1
only possible cases of f(k)=1 is k=0 and (p=3,k=1)
So, it is obvious from this, that w=1,3 are the only possible cases
Edited on July 13, 2007, 12:26 pm
Edited on July 13, 2007, 12:29 pm
Solution
