How to show that $$(a^{n},b^{n})=(a,b)^{n}$$ and $$[a^{n},b^{n}]=[a,b]^{n}$$ without using modular arithmetic? Seems to have very interesting applications.$$$$Try: $(a^{n},b^{n})=d\Longrightarrow d\mid a^{n}$ and $d\mid b^n$
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1What is the second notation? For the first one, can you consider the prime factorization of $a$ and $b$? – zarathustra Aug 05 '13 at 13:12
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1@Antoine, $[a, b]$ stands for least common multiple of $a$ and $b$. – Andreas Caranti Aug 05 '13 at 13:13
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@Antoine That's right. – benjamin_ee Aug 05 '13 at 13:20
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Where is this problem from? Are you taking a course (if so which), or reading a book (if so which), or a friend told you, or what? If we know more about your mathematical background (e.g. do you know the fundamental theorem of arithmetic, about prime factorization) and the context of the problem (e.g. is it homework), we can give an answer that's at an appropriate level for you. – ShreevatsaR Aug 06 '13 at 16:28
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This guy clearly just wants a detailed solution to copy for his homework, please do not give it to him.. – sxd Aug 06 '13 at 16:29
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If $p$ is a prime and $p^t$ is the highest power of $p$ dividing $a$, then $p^{tn}$ is the highest power dividing $a^n$.
Therefore $\text{gcd}(a^n,b^n)=\text{gcd}(a,b)^n$.
For the other one, start with $$ \text{lcm}(a,b)=\frac{ab}{\text{gcd}(a,b)} $$ and take $n$-th powers both sides.
Andrea Mori
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@marcelolpjunior: sorry, I don't know how to make it simpler than this. – Andrea Mori Aug 06 '13 at 12:45
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Bezout's Identity says that $\gcd(a,b)$ is the smallest positive element of $\{ax+by:x,y\in\mathbb{Z}\}$.
The smallest positive element of $\{acx+bcy:x,y\in\mathbb{Z}\}$ is $\gcd(ac,bc)$; it is also $c$ times the smallest positive element of $\{ax+by:x,y\in\mathbb{Z}\}$ which is $\gcd(a,b)$. Therefore $$ \gcd(ac,bc)=c\gcd(a,b)\tag{1} $$ Suppose $ax+by=1$ and $au+cv=1$, then $$ \begin{align} by\,cv&=(1-ax)(1-au)\\ &=1-a(x+u-axu)\\ a(x+u-axu)+bc\,vy&=1 \end{align} $$ Thus, $$ \gcd(a,b)=1\quad\text{and}\quad\gcd(a,c)=1\implies\gcd(a,bc)=1\tag{2} $$ Using $(2)$ and induction, we get that $$ \gcd(a,b)=1\implies\gcd\left(a^n,b^n\right)=1\tag{3} $$ Using $(1)$ and $(3)$, we have $$ \begin{align} \gcd(a,b)\gcd\left(\frac{a}{\gcd(a,b)},\frac{b}{\gcd(a,b)}\right)&=\gcd(a,b)&&(1)\\ \gcd\left(\frac{a}{\gcd(a,b)},\frac{b}{\gcd(a,b)}\right)&=1&&\text{cancel}\\ \gcd\left(\frac{a^n}{\gcd(a,b)^n},\frac{b^n}{\gcd(a,b)^n}\right)&=1&&(3)\\ \gcd(a,b)^n\gcd\left(\frac{a^n}{\gcd(a,b)^n},\frac{b^n}{\gcd(a,b)^n}\right)&=\gcd(a,b)^n&&\text{multiply}\\[9pt] \gcd\left(a^n,b^n\right)&=\gcd(a,b)^n&&(1)\tag{4} \end{align} $$
In this answer, it is shown that $$ \gcd(a,b)\,\mathrm{lcm}(a,b)=ab\tag{5} $$ Therefore, $$ \begin{align} \mathrm{lcm}(a,b)^n &=\frac{(ab)^n}{\gcd(a,b)^n}&&(5)\\ &=\frac{a^n\,b^n}{\gcd(a^n,b^n)}&&(4)\\[4pt] &=\mathrm{lcm}(a^n,b^n)&&(5)\tag{6} \end{align} $$
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Show that the common divisors of $a^n$ and $b^n$ all divide $(a,b)^n$ and that any divisor of $(a,b)^n$ divides $a^n$ and $b^n$ (the proofs are pretty straight forward). It might be useful to consider prime factorization for the second direction.
Similarly, show that $[a,b]^n$ is a multiple of $a^n$ and $b^n$ but that it is also the smallest such multiple (and again there, prime factorization might be useful in the second case). If you need more details just ask.
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Let's use prime factorization, I think this way everything makes more sense. To show equality, it suffices to show that the prime powers dividing either side are the same. Let's show $(a^n, b^n) = (a,b)^n$ first.
Since $p$ is prime, $p$ divides $a$ if and only if it divides $a^n$ and similarly for $b$ ; so if $p$ does not divide $a$ or $b$, then $p^0 = 1$ is the greatest power of $p$ that divides both sides. If $p$ divides both $a$ and $b$, let $p^k$ be the greatest power of $p$ dividing $(a,b)$, so that $p^{kn}$ is the greatest power of $p$ dividing $(a,b)^n$. Since $p^k$ divides $a$, $p^{kn}$ divides $a^n$, and similarly for $b$. For obvious reasons the greatest power of $p$ dividing both $a^n$ and $b^n$ must be a power of $p^n$. But if $p^{(k+1)n}$ divided both $a^n$ and $b^n$, then $p^{(k+1)}$ would divide $a$ and $b$, contradicting the fact that $p^k$ is the greatest power of $p$ dividing $(a,b)$. Therefore $p^{kn}$ is the greatest power of $p$ dividing $(a^n,b^n)$ and the greatest power of $p$ dividing $(a,b)^n$, so taking the product over all primes, $(a^n,b^n) = (a,b)^n$.
For $[a^n,b^n] = [a,b]^n$ you can do very similar techniques as with the gcd, except all the 'greatest' are replaced by 'smallest' in the last proof, and 'division' is replaced by 'being a multiple of'.
Hope that helps,
Patrick Da Silva
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@marcelolpjunior : It should've looked something like this : $(a,b)$ is by definition the greatest common divisor of $a$ and $b$. Since $(a,b)$ divides $a$ and $b$, $(a,b)^n$ divides $a^n$ and $b^n$, so this means $(a,b)^n$ divides $(a^n, b^n)$. In the other direction, if you look at a power of $p$ prime dividing $(a^n,b^n)$, then write $p^t = p^{kn+l}$ using the Euclidean algorithm, and it follows that $p^{kn+l}$ divides $a^n$ and $b^n$, so we must have $p^{kn}$ dividing $a^n$ and $b^n$ also, so that $p^k$ is smaller than the biggest power of $p$ dividing $(a,b)$. – Patrick Da Silva Aug 05 '13 at 14:23
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If $p^k$ is equal to that power, we must have $l = 0$ by taking the $n^{th}$ power. This shows that $(a^n,b^n)$ also divides $(a,b)^n$. – Patrick Da Silva Aug 05 '13 at 14:24
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let gcd(a,b)=d
$\Longrightarrow d|a \wedge d|b$
$\Longrightarrow$ $d|a^n$ $\wedge$ $d|b^n$
$\Longrightarrow$ d|$(a^n,b^n)$---------------(1)
now, $d|(a,b)$ [in fact, $(a,b)=d$]
it implies $d|(a,b)^n$-----------------(2)
by equation (1)and(2), it follows obviously that $(a^n,b^n)=(a,b)^n$
now, it is easy work to derive similar result for lcm by this method.
user89071
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This is not a proof: you have shown that $d|(a,b)^n$, but you actually need $d^n|(a,b)^n$. – ShreevatsaR Aug 06 '13 at 16:34
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1Can you elaborate on the " it follows obviously" part? Because it's not obvious to me. (And in fact it doesn't follow.) – ShreevatsaR Aug 07 '13 at 13:17
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Yes I still don't understand, even though I have indeed studied the definition and properties carefully. Please tell me. :-) In general, whenever in a proof you write "it follows obviously" or "clearly" or so on, that is the usual place errors are found. – ShreevatsaR Aug 08 '13 at 13:48
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