Greatest common divisor of powers of $a,b$

Question

For all integers $a, b$ and nonnegative integers $m, n$, we have $\gcd(a^m, b^n ) = \gcd(a, b)^{\min(m,n)}$.

I have no idea where to go with this problem and how to even start it.

Answer 1

Here is a set up: Let $gcd(a,b)=x$, then we can write $a=px$ and $b=qx$. Now what is the gcd of $p$ and $q$? All right, move on. Let $a^m=p^mx^m$ and $b^n=q^nx^n$. Is it obvious what $p^m$ and $q^n$ have in common, which is not a whole lot? Ok, now for sure $x^m$ and $x^n$ do have something in common, right? Why would we be interested in the lowest exponent of the two? Because that answer will help you in the right hand side of the equation you need to prove.

Answer 2

It is not true. Take $a=8=2^3, b=4=2^2, m=2, n=3$. Then $a^m=b^n=2^6$, so $\gcd(a^m,b^n)=2^6$, but $\gcd(a,b)^{\min(m,n)}=(2^2)^2=2^4$

Answer 3

As stated by Ross Millikan, this is not true.

What is true is that if $a = \prod_i {p_i^{\alpha_1}}$ and $b= \prod_i {p_i^{\beta_1}}$ are the prime factorisation of a and b, then for $n,m \in \Bbb N$ :

$$gcd(a^n, b^m) = \prod_{i=1}^{+\infty} {p_i ^{\min (n \alpha_i, m \beta_i)}}$$

Thus your property only holds iff for $i \in \Bbb N$, you have $\alpha_i = 0$ or $\beta_i = 0$ or $\alpha_i = \beta_i$ (if I am not mistaken).

That is iff $a\over{gcd(a,b)}$ and $gcd(a,b)$ are relatively prime and $b\over{gcd(a,b)}$ and $gcd(a,b)$ are relatively prime.