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?

let A=p1^k1 * ... *pn^kn  with p1,..,pn all prime and k1,..,kn>=0

now the solutions to xyz=A can be shown as

x=p1^a1*...*pn^an
y=p1^b1*...*pn^bn
z=p1^c1*...*pn^cn
where at+bt+ct=kt for all t

now kt can be represented as the sum of 3 positive integers in (kt+1)(kt+2)/2 ways

let F(x)=(x+1)(x+2)/2 then D(A) can be written as

Product of F(kt) for t=1 to n

and d(A) is equal to the product of (kt+1) for t=1 to n

D(A)^2-2D(A)+d(A) is then equal to the product of this whole thing for t=1 to n

(kt+1)^2(kt+2)^2/4 - 2(kt+1) + (kt+1)

now we can simplify that expression like so

[(kt+1)^2(kt+2)^2-4(kt+1)]/4
[(kt+1)( (kt+1)(kt+2)^2-4) )]/4
[(kt+1)((kt+1)(kt^2+4kt+4)-4)]/4
[(kt+1)(kt^3+5kt^2+8kt)]/4
[kt(kt+1)(kt^2+5kt+8)]/4

now since kt>=0 for all t then obviously each of these terms is positive and thus

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

now for equality all of those terms must equal zero

thus kt(kt+1)(kt^2+5kt+8)=0  or
kt=0 kt=-1 kt=(1/2)(-5+-i*Sqrt(7))
now the only value that fits with kt>=0 is kt=0 thus A=1

if A=1 then D(A)=1 and d(A)=1  and D(A)^2-2D(A)+d(A)=1-2+1=0

QED

Posted by Daniel on 2007-08-13 01:57:05