Let $a,b$ be natural numbers. Show that $gcd(a^n,b^n)$ = ($gcd(a,b)^n)$ for any integer $n$.

How I started was first proof by contradiction, and then tried to do an inductive proof when that didn't work, but neither of them worked out for me. I think I'm confused as to the notation I should use. Hmmm.

I'm not looking for someone to answer the question fully, I just need a nudge in the right direction.


Okay, after the super helpful hints, here is what I have for the proof:

Suppose $gcd(a,b)$ = $c$ for some integer $c$

Then $c$ = ${p_1}^{min\{a_1, b_1\}\ }$... ${p_k}^{min\{a_k, b_k\}}$ for some positive integer $k$

So $gcd(a^n, b^n)$ = ${p_1}^{min\{na_1, nb_1\}\ }$... ${p_k}^{min\{na_k, nb_k\}}$ = ${p_1}^{nmin\{a_1, b_1\}\ }$... ${p_k}^{nmin\{a_k, b_k\}}$

Now we see ($gcd(a, b)^n$) = $(c)^n$ = (${p_1}^{min\{a_1, b_1\}\ }$... ${p_k}^{min\{a_k, b_k\}})^n$ = ${p_1}^{nmin\{a_1, b_1\}\ }$... ${p_k}^{nmin\{a_k, b_k\}}$

Therefore, $gcd(a^n, b^n)$ = $(gcd(a, b)^n)$ for all positive integers $n$

Is this too short? Should I show the prime factorization of $a$ and $b$ before showing the prime factorization of $gcd(a, b)$ ?

  • $\begingroup$ Hint: go prime by prime. Suppose $p^k$ is the largest power of p that divides gcd(a,b). Then $p^{nk}$ is the largest power of p dividing the right hand of your conjectured equation. Could the power of p be greater on the left? Could it be less? $\endgroup$ – lulu Jul 8 '15 at 21:52
  • $\begingroup$ See this answer for a few proofs of this GCD Freshman's Dream. $\endgroup$ – Gone Jul 8 '15 at 22:16

If you're just looking for a nudge in the right direction try looking at the prime factorization of $a$ and $b$ vs $a^n$ and $b^n$. (You could also try thinking of $a$ as $gcd(a,b)*p_1p_2....$ and $b$ as $gcd(a,b)*q_1q_2....$ where $p_i$ and $q_j$ are all prime). Further "nudge": Could it possibly be the case that $p_i=q_j$ for any $i,j$ pair?

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A start: Let $d=\gcd(a,b)$, and let $a=da_1$ and $b=db_1$. We want to show that $\gcd(a^n,b^n)=d^n$, or equivalently that $\gcd(a_1^n,b_1^n)=1$.

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