Let A be a positive integer. Let D(A) be the number of solutions of xyz=A where x, y, and z are all positive integers; thus, D(6)=9. Let d(A) be the number of positive divisors of A, so d(6)=4.

Show that d(A)²-2D(A)+d(A)≥0. When does the equality hold?

(In reply to re: solution with proof by Praneeth)

ok, my bad.  Well that does not change my equations for D(A) and d(A)  so to recap

A=p1^k1*p2^k2*...*pn^kn  being the prime factorization of A

D(A)=Product[(kt+1)(kt+2)/2,t=1 to n]

d(A)=Product[(kt+1),t=1 to n]

I now realize I made another serious error in my proof and so now I shall take a different approach

first let us make note of the fact that D(A) can be written in terms of d(A) like so

D(A)=Product[(kt+1)(kt+2)/2,t=1 to n]

D(A)=Product[(kt+1),t=1 to n] * Product[(kt+2)/2,t=1 to n]

D(A)= d(A) * Product[(kt/2)+1,t=1 to n]

now note that every term of the product that remains is greater than 1, thus the entire product is greater than one and therfore we can state that

D(A)=v*d(A)   where v>=1

so now we can rewrite d(A)^2-2D(A)+d(A) as

d(A)^2-2*v*d(A)+d(A)

d(A)*(d(A)-2*v+1)

now obviously d(A)>2v and thus d(A)*(d(A)-2*v+1)>=0 and thus

d(A)^2-2D(A)+d(A)>=0

now for equality we simply need that

d(A)=2v-1 or

Product[kt+1,t=1 to n]=2*Product[(kt+2)/2,t=1 to n]-1

2^n*Product[kt+1,t=1 to n]=2*Product[kt+2,t=1 to n]-2^n

now this reduces down to

2*(2^(n-1)-1)*Product[kt,t=1 to n]=0

so we either have 2^(n-1)=1  or n=1 implying that all prime powers give equality

or we have kt=0 for all t  thus giving A=1

so in conclusion the values of A for which

d(A)^2-2D(A)+d(A)=0 are

A=1 and A=p^n with p a prime and n>0

 

Posted by Daniel on 2007-08-14 02:07:53