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?
For any positive integer i,
d(i²)≥2d(i)-1
sum all the terms for every i which satisfies i|A
=> Σd(i²)≥2Σd(i) - Σ1
=> d(A)²≥2D(A)-d(A)
=> d(A)²-2D(A)+d(A)≥0
Hence proved.
No. of solutions of solutions of xyz=A is Σd(i) for every
i>0 which satisfies i|A.
Σd(i²) = π(Σ(2j+1)) 0 ≤ j ≤ (qk+1)
where qk: index of kth prime divisor of A.
Σd(i²) = π(qk+1)²= d(A)².
Edited on July 23, 2008, 8:02 am
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Posted by Praneeth
on 2008-07-23 08:01:12 |
Other Simple Solution
