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?
Consider a number x which is a divisor of A. Let P(x,A) be no. of positive divisors of A not greater than x. All divisors of x are less than or equal to x and also factors of A. So, P(x,A)>=d(x).--(1)
D(A)=Σ{No. of solutions of xy=i}; where i takes all possible values of factors of A.
D(A)=Σd(i) <= ΣP(i,A) (from (1))
P(i,A) varies from 1 to d(A) when i varies from 1 to A.
D(A) <= 1+2+3+....+d(A) <= (1/2)*d(A)*(d(A)+1).
Hence, d(A)²-2D(A)+d(A)≥0.
The equality holds when d(i)=P(i,A); for every i which is a divisor of A.--(2)
Let p2>p1 are two distinct prime divisors of A, then
P(p2,A)>=d(p2)+1(This is because p1 is a prime divisor of A less than p2 and by definition of P(x,A)).
So, there can't be two or more distinct prime factors for A for the equality to hold.(from eq(2))
If n(A): no. of distinct prime divisors of A, then equality holds for A whose n(A)= 0 or 1.
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