$(a^{n},b^{n})=(a,b)^{n}$ and $[a^{n},b^{n}]=[a,b]^{n}$?

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 $d\mid b^n$

Solution 1:

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.

Solution 2:

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, $$ \begin{align} \mathrm{lcm}(a,b)^n &=\frac{(ab)^n}{\gcd(a,b)^n}&&(5)\\ &=\frac{a^n\,b^n}{\gcd(a^n,b^n)}&&(4)\\[4pt] &=\mathrm{lcm}(a^n,b^n)&&(5)\tag{6} \end{align} $$