Exercise 1.3.7 ($n$ is a product of square-free and square)

Proof. Write $n=p_1^{a_1}{p}_2^{a_2}\cdots p_l^{a_l}$, the decomposition of $n$ in prime factors, where $a_i>0$ for each index $i$. The division of $a_i$ by $2$ gives $a_i=2b_i+r_i$, where $r_i=0$ or $r_i=1$. Then $n=\mathit{ab}$, where

Since $r_i\in \{0,1\}$ for each $i$, $a$ is square-free, and $b$ is a square. This shows that $n$ can be written in the form $n=\mathit{ab}$, where $a$ is square-free and $b$ is a square.

Now we prove the unicity of such a decomposition. Suppose that $n=\mathit{ab}=\mathit{cd}$, where $a,c$ are square-free and $b,d$ are squares. Let $q_1,q_2,\dots ,q_u$ the prime factors of $a,b,c,d$. Since $b,d$ are squares, all the exponents in the decompositions of $b$ and $d$ are even.

Write

where $s_i\ge 0,t_i\ge 0,c_i\ge 0,d_i\ge 0$ for all $i\in [[1,u]]$. Since $a,c$ are square-free, $s_i\in \{0,1\},t_i\in \{0,1\}$ for all $i$.

Then $\mathit{ab}=\mathit{cd}$ gives

The unicity of the decomposition in prime factors shows that

Since $0\le s_i<2,0\le t_i\le 2$, the unicity of the pair quotient-remainder in the Euclidean division gives $s_i=t_i,c_i=d_i$ for all $i$. Thus $a=c$ and $b=d$, and the decomposition is unique. □