If $a,b,c \in \mathbb{Z}$ with $gcd(a,b)=1$ where $ab=c^n$, then $\exists d,e\in \mathbb{N}$, such that $a=d^n$ and $b=e^n$

Question

If $a,b,c \in \mathbb{Z}$ with $gcd(a,b)=1$ where $ab=c^n$, then $\exists d,e\in \mathbb{N}$, such that $a=d^n$ and $b=e^n$

I was not exactly sure how to start, but I decided to write out prime factorizations to complete the proof. I think my proof does not make much sense but this is what I have so far. Would appreciate some input.

Proof: Since $a$ and $b$ are relatively prime, they have have unique disjoint prime factorizations:
$$a=p_1^{a_1} \cdot p_2^{a_2}\cdots p_s^{a_s}\\
b=q_1^{b_1} \cdot q_2^{b_2} \cdots q_t^{b_t}$$

Now, let us assume that $a\cdot b=c^n$, that is,
$$ab=\left(p_1^{a_1} \cdot p_2^{a_2}\cdots p_s^{a_s}\right)\left(q_1^{b_1} \cdot q_2^{b_2} \cdots q_t^{b_t} \right)=c^n$$

Since the sets $\{p_1,\ldots,p_s\}$ and $\{q_1,\ldots,q_t\}$ are disjoint, it must be the case that:
$$ab=\underbrace{\left(p_1^{a_1} \cdot p_2^{a_2}\cdots p_s^{a_s}\right)}_{d^n}
\underbrace{\left(q_1^{b_1} \cdot q_2^{b_2} \cdots q_t^{b_t} \right)}_{e^n}=c^n$$
Hence, $a=d^n$ and $b=e^n$.

Answer 1

Yes, the proof is correct. Let $c=\Pi_{i=1}^k {r_i}^{e_i}$ (prime factorization of c) Then, think of ${r_i}$ for all integers $i$ ($1 \le i \le k$)

Note that $\gcd(a,b)=1$. So,note that if $r_i | a$ then $\gcd(b,r_i)=1$, and vice versa. ...(1)

$a \times b = c^n = \Pi_{i=1}^k {r_i}^{n \times e_i}$

Let $P= \{ p_i|1 \le i \le s, i \in \boldsymbol Z \}$, $Q= \{ q_i|1 \le i \le t, i \in \boldsymbol Z \}$, $R= \{ r_i|1 \le i \le k, i \in \boldsymbol Z \}$

then, $P \cup Q = R$ and $ P \cap Q = \varnothing$

Let $v_p(n)$ be the exponent of the largest power of p that divides n.

Then, $n|v_{r_i}(c^n)$ for all integers $i$ ($1 \le i \le k$) ...(2)

by (1),(2),

$n|v_{p_i}(a)$ for all integers $i$ ($1 \le i \le s$) and

$n|v_{q_i}(b)$ for all integers $i$ ($1 \le i \le t$)

So the proof is correct.