# Divisibility (gcd) 15

• December 20th 2008, 11:18 PM
Sea
Divisibility (gcd) 15
Show that:

$n \in \mathbb{N},$

$(a,b)=1 \Rightarrow (a^{n},b^{n})=1$
• December 21st 2008, 12:55 AM
o_O
Let the prime factoirzations be: $a = p_1^{e_1}p_2^{e_2}\cdots p_n^{e_n}$ and $b = q_1^{f_1}q_2^{f_2}\cdots q_m^{f_m}$ .

Since $(a,b)=1$, there are no common factors between them, i.e. $p_i \neq q_j$ for any $1 \leq i \leq n$, $1 \leq j \leq m$.

What can you say about $a^n$ and $b^n$?
• December 21st 2008, 01:18 AM
Sea
$a^n = (p_1^{e_1})^{n}(p_2^{e_2})^{n}\cdots (p_n^{e_n})^{n}$

$b ^n= (q_1^{f_1})^{n}(q_2^{f_2})^{n}\cdots (q_m^{f_m})^{n}$

and

I think...

$(a,b)=(p_1^{e_1}p_2^{e_2}\cdots p_n^{e_n},q_1^{f_1}q_2^{f_2}\cdots q_m^{f_m})=1 \Rightarrow (p_i,q_j)=1, \hspace{0.3cm}p_i \neq q_j ,\hspace{0.3cm}1 \leq i \leq n$, $1 \leq j \leq m$.
• December 21st 2008, 03:19 AM
PaulRS
Suppose $(a^n,b^n)>1$ then there is a prime $p$ such that $p|a^n$ and $p|b^n$

You can check by Euclid's Lemma that we must have $p|a$ and $p|b$ thus $(a,b)\geq{p}>1$ CONTRADICTION!
• December 21st 2008, 03:55 AM
Sea
I understand...:)...

Thanks a million...;):)...