If a^2 divides b^2, a divides b

[DerpyPenguin]

Homework Statement

If a^2 divides b^2, then a divides b
Also
If a^2 divides b^3, then a divides b

Homework Equations

The Attempt at a Solution

For the first question, if a^2 divides b^2, then b^2=(a^2)c where c is some integer
c=(b^2)/(a^2)
c=(b/a)^2
The answers in the back say that it is enough to show that c is a perfect square, but I don't see how that's sufficient. If c is a perfect square, then c is an integer, which implies that a^2 divides b^2, which we already knew.

No idea where to go for the second question. The same strategy doesn't seem to work.

 

[Fightfish]

It might help to try to establish this fact: the square root of an integer is either an integer or an irrational number.

 

[PeroK]

Homework Statement

If a^2 divides b^2, then a divides b
Also
If a^2 divides b^3, then a divides b

Homework Equations

The Attempt at a Solution

For the first question, if a^2 divides b^2, then b^2=(a^2)c where c is some integer
c=(b^2)/(a^2)
c=(b/a)^2
The answers in the back say that it is enough to show that c is a perfect square, but I don't see how that's sufficient. If c is a perfect square, then c is an integer, which implies that a^2 divides b^2, which we already knew.

No idea where to go for the second question. The same strategy doesn't seem to work.

A more general approach to those problems would start with:

Let $p$ be a prime divisor of $a$ of order $n$.

 

[PeroK]

No idea where to go for the second question. The same strategy doesn't seem to work.

You could try to find a counterexample!