1

Show that if $a$ and $b$ are positive integers with $(a,b)=1$ then $(a^n, b^n) = 1$ for all positive integers n

Hi everyone, for the proof to the above question,

Can I assume that since $(a, b) = 1$, then in the prime-power factorization of a and b, they have no prime factor in common, when they are taken to the $nth$ power, they will still have no prime factors in common, and so $(a^n, b^n) = 1$ for all positive integers n.

I think I'm jumping to conclusions here again, if so, leave some tips on how to do the proof properly, thanks :)

And also, I do not know how to approach the reverse problem where if $(a^n, b^n) = 1 then (a, b) = 1$, any guidance will be much appreciated!

DJ_
  • 1,400
  • 1
    Prime factorization is exactly the right idea. Do it for a and b, and see what happens when you raise them to the $n^{th}$ power for the first part. Then do the same argument in reverse. –  Oct 23 '13 at 06:12
  • Yes I wrote when raised to the nth power there are still no common prime divisors however can I jump to the assumption thus $a^n and b^n$ are relatively prime as well – DJ_ Oct 23 '13 at 06:16
  • You can argue by contradiction as well : If $p\mid a^n$ and $p\mid b^n$ where $p$ is prime, then $p\mid a$ and $p\mid b$ (why?). – Prahlad Vaidyanathan Oct 23 '13 at 06:23
  • 1
    The reverse problem follows immediately from the Bézout identity. If $a^n x + b^n y = 1$ then $a (a^{n-1}x) + b (b^{n-1}y) = 1$ and so $a,b$ are coprime. – copper.hat Oct 23 '13 at 06:37

2 Answers2

9

Here's an approach that doesn't use prime factorization.

Lemma: For all $m,r,k\in \Bbb N$, $\gcd(m,k)=1\implies \gcd(m,k^r)=1$.

Proof: Let $m,r,k\in \Bbb N$ be such that $\gcd(m,k)=1$.

Bézout yields $um+vk=1$, for some $u,v\in \Bbb Z$.

Thus $$1=1^r=(um+vk)^r=\sum \limits_{j=0}^r\left({r\choose j}(um)^{r-j}(vk)^j\right)=\sum \limits_{j=0}^{r-1}\left({r\choose j}(um)^{r-j}(vk)^j\right)+(vk)^r,$$ which implies $$m\underbrace{\color{blue}{\sum \limits_{j=0}^r\left({r\choose j}u^{r-j}m^{r-j-1}(vk)^j\right)}}_{\huge \in \Bbb Z}+k^r\color{blue}{v^r}=1.$$

Again Bézout says that $\gcd(m,k^r)=1$ (due to the blue scalars).$\,\square$

You can now use the lemma twice: $\gcd(a,b)=1\implies \gcd(a,b^n)=1\implies \gcd(a^n,b^n)=1$.

Git Gud
  • 31,356
0

For those type of proofs, assuming $ gcd(a^n,b^n) \neq 1 $ and showing a contradiction is a good way.

Assume $ gcd(a^n,b^n) \neq 1 $. Then $\exists d_1$ such that $d_1|a^n \land d_1|b^n \land d_1>1$. This implies:

Let's check out what $\exists d_1$ such that $d_1|a^n \land d_1>1$ implies to.

$ d_1|a^n \implies (d_1|a \,\,\vee d_1=a^m)$ where $ m|n$.

Samely, $ d_1|b^n \implies (d_1|b \,\,\vee d_1=b^k)$ where $ k|n$.

Then, we have:

$(d_1|a \,\,\vee d_1=a^m) \land (d_1|b \,\,\vee d_1=b^k) \land m|n \land k|n \land d_1>1$.

We know that there is no $d_1>1$ satisfying $d_1|a \land d_1|b$, because we know that $gcd(a,b)=1$.

On the other hand, as the proposition should be satisfied for all $n$, if we selected $n$ as a prime number, we wouldn't be able to find such $m\neq1$ and $k\neq1$. Then $m=n=1$, that is $(d_1=a \land d_1|b) \vee (d_1=b \land d_1|a)$. Then $gcd(a,b)=a \vee gcd(a,b)=b$ which is a contradiction.

Then our assumption is false.

Zafer Sernikli
  • 881
  • 4
  • 11