Greatest common divisor of a number and the same number multiple of some rational

Question

I want to simplify (e.g. in terms of prime factors and its exponents) given expression: $$ \gcd\left(a, a \frac{b}{c}\right), $$ where $c\mid ab$.

Is it possible?

Thanks!

Answer 1

Conside the prime factorisation:

$$a = \prod\limits_i p_i^{k_i}, b=\prod\limits_i p_i^{l_i}, c=\prod\limits_i p_i^{m_i}$$

Then

$$\begin{align} \gcd\left(a, a \dfrac{b}{c}\right) & = \gcd\left(\prod\limits_i p_i^{k_i}, \prod\limits_i p_i^{k_i+l_i-m_i}\right)\\ & =\prod\limits_i p_i^{k_i-m_i}\gcd\left(\prod\limits_i p_i^{m_i}, \prod\limits_i p_i^{l_i}\right)\\ & = \dfrac{a \gcd(b,c)}{c} \end{align}$$

Answer 2

Not really. For example, if $c\mid b$ and $\gcd(a,b) = 1$, then your expression is equal to $a$; on the other hand, if $c\mid a$ and $\gcd(c,b) = 1$, then the expression is $\frac{a}{c}$.

Answer 3

Let $a=\prod\limits_{i =0}^kp_i^{r_i},b=\prod\limits_{i =0}^kp_i^{s_i},c=\prod\limits_{i =0}^kp_i^{t_i}$, then we have

$$\frac {ab}c=\prod_{i =0}^kp_i^{r_i+s_i-t_i}\\ \text{gcd}\left(a,\frac{ab}c\right)=\prod_{i=0}^kp_i^{\min(r_i,r_i+s_i-t_i)}$$

This does not simplify as I initially thought to gcd$\left(a,\frac {ab}c\right)=\frac a{\text{gcd}(a,c)}$ but instead includes the factors present in $b$, probably something like

$$\frac{a\cdot\text{gcd}(a,b,c)}{\text{gcd}(a,c)}$$

where gcd$(a,b,c)=\text{gcd}(a,\text{gcd}(b,c))$.

The short answer is no, the expression does not really simplify from your starting point.