I need to prove that the product of two numbers equals the product of their gcd and lcm. I cant prove it. it's just classic number theory, but it's hard. any help??
 A: Here is one proof. First note the following:
$$\alpha+\beta = \min\{\alpha,\beta\} + \max\{\alpha,\beta\}$$
Let us look at the prime decomposition of $a$ and $b$. $$a = \prod_{k=1}^{\infty}p_k^{\alpha_k} \text{ and }b = \prod_{k=1}^{\infty}p_k^{\beta_k}$$
where $\alpha_k, \beta_k \in \mathbb{Z}^+ \cup \{0\}$. Though the product is an infinite product, beyond some $k$, we will have $\alpha_k, \beta_k = 0$ and hence we are only multiplying $1$'s. We then have $$\gcd(a,b) = \prod_{k=1}^{\infty}p_k^{\min\{\alpha_k, \beta_k\}} \text{ and }\text{lcm}(a,b) = \prod_{k=1}^{\infty}p_k^{\max\{\alpha_k, \beta_k\}}$$
Hence,
$$\gcd(a,b) \cdot \text{lcm}(a,b) = \prod_{k=1}^{\infty}p_k^{\min\{\alpha_k, \beta_k\} + \max\{\alpha_k, \beta_k\}} = \prod_{k=1}^{\infty}p_k^{\alpha_k + \beta_k} = \prod_{k=1}^{\infty}p_k^{\alpha_k} \cdot \prod_{k=1}^{\infty}p_k^{\beta_k} = a \cdot b$$
A: Suppose your numbers are $a\times c$ and $b\times c$, with $c$ the common factor and $a$ and $b$ relative prime. From there it should be straight-forward.
A: Prove that for each maximal prime power $p^k$ dividing $ab$, that the same power of $p$ is the maximal power of $p$ dividing $\gcd(a,b)\operatorname{lcm}(a,b)$. That's one roundabout way to show two integers are equal.
To do that, note that if $p^m$ is the maximal power of $p$ dividing $a$, and $p^n$ is the maximal power of $p$ dividing $b$, then $p^{\min(m,n)}$ is the maximal power of $p$ dividing $\gcd(a,b)$ and $p^{\max(m,n)}$ is the maximal power of $p$ dividing $\operatorname{lcm}(a,b)$. Now you need to show that $m+n=\min(m,n)+\max(m,n)$, which should be clear.
