Divisibility (gcd) 15

**Sea** · December 20th 2008, 11:18 PM

Show that:

$n \in \mathbb{N},$

$(a,b)=1 \Rightarrow (a^{n},b^{n})=1$

**o_O** · December 21st 2008, 12:55 AM

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 $b^n$ ?

**Sea** · December 21st 2008, 01:18 AM

$a^n = (p_1^{e_1})^{n}(p_2^{e_2})^{n}\cdots (p_n^{e_n})^{n}$

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

and

I think...

$(a,b)=(p_1^{e_1}p_2^{e_2}\cdots p_n^{e_n},q_1^{f_1}q_2^{f_2}\cdots q_m^{f_m})=1 \Rightarrow (p_i,q_j)=1, \hspace{0.3cm}p_i \neq q_j ,\hspace{0.3cm}1 \leq i \leq n$ , $1 \leq j \leq m$ .

**PaulRS** · December 21st 2008, 03:19 AM

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!

**Sea** · December 21st 2008, 03:55 AM

I understand...

...

Thanks a million...

...

Thread: Divisibility (gcd) 15

LinkBack

Thread Tools

Display

Divisibility (gcd) 15

Similar Math Help Forum Discussions

Divisibility 11

Divisibility (gcd) 10

Divisibility (gcd) 9

Divisibility (gcd) 8

Divisibility

Search Tags