Proving Divisibility: How to Show b|a When b3|a2 - Helpful Tips"

[audiowize]

Hello,

If we are given that b³|a², how do we show that b|a?

I started off looking at prime factorizations, but I could use a push in a more substantial direction.

 

[hamster143]

Prime factorizations are one way to go.

Or you could observe that b^3 | a^3.

 

[roam]

I know that since $$b^3 | a^3$$

Then $$a^3 = mb^3$$ ( $$\exists m \in \mathbb{Z}$$). And $$a^3 \equiv b^3 \pmod m$$.

But I don't know where to go from here...

 

[hamster143]

I know that since $$b^3 | a^3$$

Then $$a^3 = mb^3$$ ( $$\exists m \in \mathbb{Z}$$). And $$a^3 \equiv b^3 \pmod m$$.

But I don't know where to go from here...

Can you prove that m is a cube?

 

[roam]

Can you prove that m is a cube?

Hmm, that's a good idea but I'm not sure if it's possible to prove that...

 

[ramsey2879]

Hmm, that's a good idea but I'm not sure if it's possible to prove that...

If a cube is multiplied by a non-cube, is it still a cube?

 

[Bingk]

$$b^2|b^3|a^2$$ so $$b^2|a^2$$

Can you show that if $$b^2|a^2$$ then $$b|a$$

Here's how I did it:
a and b are integers such that b^2|a^2 => mb^2 = a^2
Since the square root of a^2 = a, then we can take the sqrt(mb^2) = b*sqrt(m) = a.
Since a and b are integers, sqrt(m) must also be an integer
=> b|a

If you want to see some other ways check out https://www.physicsforums.com/showthread.php?t=337837&page=2"