Determine all positive whole numbers w such that d(w) = √w, where d(w) denotes the number of positive divisors of w.

Since w has to be a square number (otherwise √w wouldn't be an integer) d(w) has to be odd.

Note: The equations are really messy in this form. Might help if you write them on paper.

Now consider d(Π((a_i)^(n_i))²)  ,1≤i≤k, a is the i:th prime of some unknown set with k elements. Since d(x) is multiplicative we get d(Π((a_i)^(n_i))²)=Πd²((a_i)^(n_i))=Π((n_i)+1)²

If we place this into the given equation we get:

Π((n_i)+1)²=Π((a_i)^(n_i)) , from this we can see that k has to be 2 since otherwise: k>2; Π((n_i)+1)²<2^(Σ(n_i)) (2 is chosen as a base since a_i is a prime and 2 is the smallest prime) and
Π((a_i)^(n_i))≥2^(Σ(n_i)) which makes a contradiction.

If k=2 we can write the equation as (r+1)²(p+1)²=(x^r)*(y^p)
,where r and p are primes and x,y are integers. Only solution satisfying this equation in integers is p=r=2 from which we get w=9. This is the only solution since (r+1)²(p+1)²≤(x^r)*(y^p) for all other integers satisfying the conditions.

Then there is the trivial case r=p=0 from which we get w=1.

So the equation is satisfied by integers 1 and 9.

Posted by atheron on 2007-07-13 12:55:01