Check out some similar questions!

mrenney · Dec 7, 2010, 04:10 PM

that if a^3 divides b^2, then a divides b. (Hint: Write a and b as their prime factorizations, a = p sub1 raised to the e sub1... p subn raised to the e subn and b = p sub1 raised to the d sub1... p subm raised to the d subm, then consider what you know about the exponents in the prime factorizations of a^3 and b^2.)

tickle · Dec 7, 2010, 04:37 PM

So you post your homework questions and then leave without really finding out if someone answered.

So someone did. We have clear guidelines on assisting with homework questions, mreeney, and you left, what three of them??

Tick

mrenney · Dec 7, 2010, 10:14 PM

I got kicked offline at my school's wifi, could you help with the problem now?

ebaines · Dec 8, 2010, 01:32 PM

You're given:

$ \frac {B^2} {A^3} = k $

where k is an integer. Thus

$ \frac {B^2} {A^2} = kA $

is also an integer. If you show A and B as the product of their primes:

$ \frac {B^2}{A^2} \ = \ \frac {(b_1b_2b_3 ...b_m)(b_1b_2b_+3...b_m)}{(a_1a_2a3 ...a_n)(a_1a_2a_3...a_n)} = kA $

Each of the prime factors in the numerator must have a corresponding prime factor in the denominator. This means that each of A's primes is present in the numerator, and you can divide through by all the a's. This leaves:

$ (b_jb_k...b_m)(b_jb_k...b_m) = kA $

Hence A divides into B:

$ \frac B A = b_jb_k...b_m $

mrenney · Dec 9, 2010, 07:56 AM

How does (bsubj bsubk... bsubm)(bsubj bsubk... bsubm) become B at the end, wouldn't it still be B^2?

ebaines · Dec 9, 2010, 10:10 AM

From the second to last line in my previous post:

$ (b_jb_k...b_m)(b_jb_k...b_m) = kA $

combined with the fact that

$ \frac {B^2}{A^2} = kA $

Tells you that

$ \frac {B^2} {A^2} = (b_jb_k...b_m)\cdot (b_jb_k...b_m) $

Take the square root of both sides:

$ \frac B A = (b_jb_k...b_m) $

Hence B is a multiple of A.