- #1

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If we are given that b

^{3}|a

^{2}, 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.

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- #1

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If we are given that b

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

- #2

- 907

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Prime factorizations are one way to go.

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

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

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- #3

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

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

- #4

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Can you prove that m is a cube?

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

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

- #5

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Hmm, that's a good idea but I'm not sure if it's possible to prove that...Can you prove that m is a cube?

- #6

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If a cube is multiplied by a non-cube, is it still a cube?Hmm, that's a good idea but I'm not sure if it's possible to prove that...

- #7

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[tex]b^2|b^3|a^2[/tex] so [tex]b^2|a^2[/tex]

Can you show that if [tex]b^2|a^2[/tex] then [tex]b|a[/tex]

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"

Can you show that if [tex]b^2|a^2[/tex] then [tex]b|a[/tex]

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"

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