Prove $ab = \text{gcd}(a,b)\text{lcm}(a,b)$

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The problem statement is as follows:

The least common multiple of $2$ nonzero integers, $a,b$, denoted $\text{lcm}(a,b)$ is the smallest positive integer n so that $$a \mid n\ \text{and}\ b\mid n$$Prove that $ab = \text{gcd}(a,b)\text{lcm}(a,b)$

Hint: Express both in terms of the prime factorizations of $a$ and $b$

Summary of my attempt/thought process below

Specific questions in last section


To start, I've written the prime factorizations as: $$a = p^{n_1}_1 \cdots p^{n_k}_k$$ $$b = q^{m_1}_1 \cdots q^{m_s}_s$$ Where $q,p$ prime and $n_i, m_j \in \mathbb{Z}_{+}$. So that, $$ab = p^{n_1}_1 \cdots p^{n_k}_k \cdot q^{m_1}_1 \cdots q^{m_s}_s$$ Which we could reindex so that for any $p_i = q_j$ we take $p_i^{n_i + m_j}$ where we take (without loss of generality) $k < s$ so that we have $$ab = p_1^{n_1 + m_1} \cdots p_k^{n_k + m_k} \cdot q_l^{m_l} \cdots q_s^{m_s}$$ Where $q_p$ are the prime factors such that $q_p \neq p_i$ for any $i,p \in \mathbb{Z}_+$

Note that $\text{gcd}(a,b)$ is the smallest positive integer of the set $\mathcal{M} = \{ma + na \mid m,n \in \mathbb{Z} \}$. Hence there exist $m_0, n_0 \in \mathbb{Z}$ such that $$\text{gcd}(a,b) = d = m_0a + n_0b = m_0(p^{n_1}_1 \cdots p^{n_k}_k) + n_0(q^{m_1}_1 \cdots q^{m_s}_s)$$

Now, the only exposure in the text we've had to the $\text{lcm}(a,b)$ is exactly the content of the problem statement. Given the properties of the $\text{lcm}(a,b) = c$ we know that $$c = m_0'a = m'_0(p^{n_1}_1 \cdots p^{n_k}_k)$$ $$c = n'_0 b = n'_0(q^{m_1}_1 \cdots q^{m_s}_s)$$

Looking up an algorithm to find $\text{lcm}(a,b)$, I discern that to find $\text{lcm}(a,b)$ one must:

  1. Write out the prime factorization of both integers in question.
  2. Count the prime factors and add each unique factor to the "list"
  3. Whenever both integers have the same prime base then consider the base raised to the highest power and add that one to the "list"
  4. Multiply each element of the list together to find the least common multiple.

I know this is a bit disorganized but I'm just trying to lay out all that I know about these concepts and trying to tie them together to prove the proposition but am not seeing the full picture. I imagine I probably don't need the algorithm in order to solve as the text should be fully contained and so all I should need to know about the least common multiple is what's said in the problem statement. I'm having trouble making this into a usable expression that when multiplied with the expression for the greatest common divisor gives us $ab$. Any help is greatly appreciated!

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Let $m=gcd(a,b)$ and $M=lcm(a,b)$. Then $a=k_1m$ and $b=k_2m$ where $k_1$ and $k_2$ are coprime. On the other hand, $M=n_1a=n_2b$ where $n_1$ and $n_2$ are as small as possible. Substituting $a=k_1m$ and $b=k_2m$ we have $$ M = n_1k_1m=n_2k_2m. $$ The condition that $n_1$ and $n_2$ be as small as possible combined with the fact that $k_1$ and $k_2$ are coprime implies $n_1=k_2$ and $n_2=k_1$. So $$ gcd(a,b)lcm(a,b)= mM = m(n_1a) = m(k_2a)=a(k_2m) = ab. $$