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For all $n \geq 1$, and positive integers $a,b$ show:

If $\gcd (a,b)=1$, then $\gcd(a^n,b^n)=1$

So, I wrote the $gcd (a,b)=1$ as a linear combination: $ax+by=1$

And, I wrote the $gcd(a^n,b^n)=1$ as a linear combination: $a^n (u)+b^n (v)=1$

can I write the second linear combinations with $x,y$ and then raise the first equation to the nth power or not?

user76568
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Lil
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  • see http://math.stackexchange.com/questions/524454/how-to-prove-greatest-common-divisor-using-bezouts-lemma – r9m Mar 02 '14 at 17:21

5 Answers5

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You can use the unique prime factorization to show this result logically . If $a$ & $b$ are relatively prime then they have no common prime factors. Therefore any power of $a$ & $b$ will just repeat each of the prime factors $n$ times. Therefore $a^n$ and $b^n$ still have no common prime factors and therefore are relatively prime.

George1811
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$$ (a,b)=1\implies ax+by=1 $$ Therefore $$ \begin{align} a^nx^n &=(1-by)^n\\ &=1+b\left(\sum_{k=1}^n\binom{n}{k}b^{k-1}(-y)^k\right) \end{align} $$ and $$ a^nx^n-b\left(\sum_{k=1}^n\binom{n}{k}b^{k-1}(-y)^k\right)=1 $$ Thus, $(a^n,b)=1$. Apply the process again to get $(a^n,b^n)=1$

Another Approach is presented in this answer.

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robjohn
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There are many ways to prove this, depending on one's background and the desired generality. Below are a few, some well-known, some not. First, the most obvious is to use the fundamental theorem of arithmetic, i.e. existence and uniqueness of prime factorizations, e.g. as follows:

Hint $\ $ prime $\,p\:|\:a^n,b^n\:\Rightarrow\:p\:|\:a,b,\:$ since prime $\:p\:|\:d_1\cdots d_k\:\Rightarrow\:p\:|\:d_1\ $ or $\,\ldots\,$ or $\:p\:|\:d_k\,$

Alternatively, it may be derived as a simple inductive consequence of Euclid's Lemma, i.e. $\,(a,b)=1=(a,\,c)\ \color{#c00}{\Rightarrow}\,(a,bc)=1.\,$ Hence for $\,c=b\,$ we infer $\,(a,b^2)=1,\,$ and by induction $\,(a,b)=1=(a,b^n)\color{#c00}\Rightarrow\,(a,b^{n+1})=1.\,$ Reversely, inducting in the same way on powers of $\,a,\,$ $\,(b^n,a) = 1\,\color{#c00}\Rightarrow\, (b^n,a^2) = 1\,\ldots\,\color{#c00}\Rightarrow\, (b^n,a^k) = 1.$

Alternatively, Gauss's Lemma (GL) yields a quick proof. Let $\rm\:{\cal C}(f)\:$ denote the content of a polynomial, i.e. the gcd of its coefficients. GL states $\rm\: {\cal C}(f\,g)\ =\ {\cal C}(f)\ {\cal C}(g)\ $ hence

$\rm\qquad\qquad\qquad\ \ 1\ =\ (a,b)\ =\ {\cal C}\:(a\ x + b)\ =\ {\cal C}\:(a\ x - b)$

$\rm\qquad\qquad \Rightarrow\ \ 1\ =\ {\cal C}\:((a\ x + b)\:(a\ x - b))\ =\ {\cal C}\:(a^2\: x^2 - b^2)\: =\: (a^2,b^2)$

Iterating shows $\rm\,(a^n,b^n) = 1\,$ for $\rm\,n = 2^k,\,$ hence for all $\rm\:n,\:$ by $\rm\,m\le n\,\Rightarrow\,(a^m,b^m)\:|\:(a^n,b^n),\,$ another example of the "up then down" (or interval) induction.

More generally, one easily proves the Freshman's Dream $\,(a,b)^n = (a,b)^n\,$ for gcds or invertible ideals, by simple high-school arithmetic using their familiar arithmetical laws (associative, commutative, distributive,cancellative) along with the more special idempotent law $(a,a) = (a).$ The Bezout-based proofs in the answer of André and Rob are special cases of this method.

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Bill Dubuque
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As you wrote, there exist integers $x$ and $y$ such that $ax+by=1$. It follows that $$(ax+by)^{2n-1}=1.$$ Expand $(ax+by)^{2n-1}$ using the Binomial Theorem. Any term $\binom{2n-1}{k}(ax)^k (by)^{2n-1-k}$ of the expansion is divisible by $a^n$ or $b^n$. For if $k\ge n$, the term is divisible by $a^n$, and if $k\le n-1$ then the term is divisible by $b^n$.

Thus $1$ is an integer linear combination of $a^n$ and $b^n$.

Remark: We used the smallest exponent $2n-1$ that makes the argument work. Any exponent $\ge 2n-1$ also works.

André Nicolas
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  • The answer, said with ideals, is: $,(1)=(a,b),\Rightarrow, (1)= (1)^{2n}! = (a,b)^{2n} \subseteq (a^n,b^n)\ \ $ – Bill Dubuque Mar 02 '14 at 18:02
  • Ah, I remember seeing this answer. I think we both have answered this question before. – robjohn Mar 02 '14 at 18:13
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    I remember answering it. It took very little time to write out an answer. Trying to find the previous answer would have taken much longer, I would probably have soon given up. – André Nicolas Mar 02 '14 at 18:18
  • That is a problem with searching this site. It is very hard to find previous instances, even when you are certain they exist. It is often far easier to simply answer again. – robjohn Mar 02 '14 at 18:19
  • Heh... actually, your answer is in response to the same question that I cite in my answer. – robjohn Mar 02 '14 at 18:23
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Lemma 1. If $a$ divides $bc$ and $\gcd(a,b)=1$, then $a|c$.

Proof. If $\gcd(a,b)=1$, then there are $x,y$ integers such that $ax+by=1$. Thus, $acx+bcy=c$, and $a$ divides $a$ and $bc$, so it divides $acx+bcy$, thus it also divides $c$.

Lemma 2. If $\gcd(e,f)=1$, then $\gcd(e,f^n)=1$ for all $n\geq 1$.

Proof (by induction on $n$). The base case $n=1$ is clear. Now suppose that if $\gcd(e,f)=1$, then $\gcd(e,f^n)=1$. Let $d$ be a common divisor of $e$ and $f^{n+1}$. Since $\gcd(e,f^n)=1$ and $d|(f^n)\cdot f$ it follows that $d$ divides $f$. But $\gcd(e,f)=1$, so $d=1$. Hence, $\gcd(e,f^{n+1})=1$ and this proves the induction step. Therefore, by the principle of mathematical induction, $\gcd(e,f)=1$ implies $\gcd(e,f^n)=1$ for all $n\geq 1$.

Proposition. If $\gcd(a,b)=1$, then $\gcd(a^m,b^n)=1$ for all $n,m\geq 1$.

Proof. By Lemma 2, if $\gcd(a,b)=1$, then $\gcd(a,b^n)=1$ for all $n\geq 1$. And again by Lemma 2, if $\gcd(b^n,a)=1$, then $\gcd(b^n,a^m)=1$ for all $m\geq 1$, as claimed.

Álvaro Lozano-Robledo
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  • That's essentially the second proof in my answer, along with a Bezout-based proof of Euclid's Lemma. – Bill Dubuque Mar 02 '14 at 19:49
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    My point was (1) you can prove something stronger (the exponents not need be the same) and (2) you do not need to do two induction arguments if you prove first the right lemma. – Álvaro Lozano-Robledo Mar 02 '14 at 20:00
  • If you read more closely, you'll see that the exponents differ in my answer too $(b^n,a^k)=1.,$ Further, "reversely, inducting in the same way" means to reuse the prior result, not reprove it, so only one induction was intended (I thought that would be clear). So they really are the same proofs. You can find stronger or more general results in the post linked in my answer. – Bill Dubuque Mar 02 '14 at 20:16
  • Your "$,(b^n,a) = 1,\color{#c00}\Rightarrow, (b^n,a^2) = 1,\ldots,\color{#c00}\Rightarrow, (b^n,a^k) = 1$" is not a 'reuse' of the prior result, but rather a second induction. In addition, I don't think we should encourage students to write proofs using arrows or "...". Even if your answer is meant to be informal, students tend to write proofs in this poor schematic way, and they may take it as validation to see it here as well. – Álvaro Lozano-Robledo Mar 02 '14 at 20:20
  • That's not how it was intended, but I can see how it might be misread that way. The answer is, by careful design, (multiple) "Hint"s, not formal proofs. We'll have to disagree on arrows, since, for example, using the universal properties of gcd and lcm greatly simplifies many proofs in number theory, and often better reveals the essence of the matter. You can see many examples in prior posts. Feedback is always welcomed, including (constructive) crtiticism. – Bill Dubuque Mar 02 '14 at 20:38