Exercise 1.8. An integer is called squarefree if it is not divisible by the square of any prime. Prove that for every $n \geq 1$ there exist uniquely determined $a>0$ and $b>0$ such that $n=a^2 b$, where $b$ is squarefree.
Proof. Suppose $n \geq 1$ and $n=p_1^{\alpha_1} \cdots p_k^{\alpha_k}$. Define
$$a=p_1^{\left\lfloor\alpha_1 / 2\right\rfloor} \cdots p_k^{\left\lfloor\alpha_k / 2\right\rfloor} \text { and } b=p_1^{\alpha_1 \bmod 2} \cdots p_k^{\alpha_k \bmod 2} \text {. }$$
We then have $n=a^2 b$ since $\alpha_i=2\left\lfloor\alpha_i / 2\right\rfloor+\left(\alpha_i \bmod 2\right)$. Moreover, $b$ is square free.
Now suppose $n=c^2 d$ for $c>0$ and $d>0$. Then $a^2 b=c^2 d$ which means $a^2 \mid c^2 d$. However, $d$ is squarefree so it follows that $a^2 \mid c^2$. Similarly $c^2 \mid a^2$, thus $\left|a^2\right|=\left|c^2\right|$. This forces $a=c$ as they are both positive. Substituting $a=c$ into $a^2 b=c^2 d$ shows $b=d$. Hence this decomposition is unique.

\begin{proof}

Exercise 1.9. For each of the following statements, either give a proof or exhibit a counter example.
(a) If $b^2 \mid n$ and $a^2 \mid n$ and $a^2 \leq b^2$, then $a \mid b$.
(b) If $b^2$ is the largest square divisor of $n$, then $a^2 \mid n$ implies $a \mid b$.
Solution.
(a) False: Let $n=36, a=2$, and $b=3$.
(b) If $n=p_1^{\alpha_1} \cdots p_k^{\alpha_k}$ and $b^2$ is the largest square divisor of $n$, then by Exercise 1.8,
$$b=p_1^{\left\lfloor\alpha_1 / 2\right\rfloor} \cdots p_k^{\left\lfloor\alpha_k / 2\right\rfloor} .$$
If $a^2 \mid n$, then $a=p_1^{\beta_1} \cdots p_k^{\beta_k}$, where $\beta_i \leq\left\lfloor\alpha_i / 2\right\rfloor$. Thus $a \mid b$.

\end{prob}

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