If $\gcd(a,b)=9,$ then what is $\gcd(a^2,b^3)\,?$

Question

I know that by the euclidean algorithm, I can obtain the following equations.

I tried some algebraic manipulation but I can't seem to determine, if $$\text{if }\;\gcd(a,b)=9,\text{ then what is }\gcd(a^2,b^3)\;?$$

Answer 1

I suggest trying many cases, with $s,t$ not divisible by $3.$ Figuring out each case should help explain what is going on.

A. $ a = 3^2 s, b = 3^2t.$

B. $a = 3^2 s, b = 3^3 t.$

C. $a = 3^2s, b = 3^4 t.$

D. $ a = 9s, b = 3^5 t. $

and so on.

BB. $a = 3^3s, b = 3^2t$

CC. $a = 3^4s, b = 3^2t.$

DD. $a = 3^5s, b = 3^2t$

and so on.

Answer 2

Note $\ (A,B) = 9\,\Rightarrow A,B = 9a,9b,\ \color{}{(a,b) = 1},\, $ so $\,\color{#c00}{(a^2,b^3) = 1}\,$ by Euler

Thus $\ (A^2,B^3) = ((9a)^2,(9b)^3) = 81(\color{#c00}{a^2},9\color{#c00}{b^3}) = 81(a^2,9) = 81(a,3)^2 = (A,27)^2$

Answer 3

$\gcd(a,b) = 9$. So $a = 9k$ and $b=9j$ and $k$ and $j$ have no factors in common.

$\gcd(a^2,b^3) = \gcd(9^2k^2, 9^3j^3)$. $9^2k^2$ and $9j^3$ have $9^2$ in common.

So $\gcd(a^2,b^3) = \gcd(9^2k^2, 9^3j^3) = 9^2*\gcd(k^2, 9j^3)$. $k, j$ have no factors in common with each other but $9$ and $k^2$ may have factors is common. If they do then that prime factor in common is $3$ and $9|k^2$.

So

CASE 1: $a = 9k$ and $b=9j$ and $3|k$. so $k= 3l$ for some $l$ and $a = 27l$. If so then $3 \not \mid j$ (as $j,k$ have no factors in common).

So $\gcd(a^2, b^3) = \gcd(27^2l^2, 9^3j^3)$. These have $9^3$ in common but $l$ and $j$ have no factors in common and no power of $3$ divides $j^3$.

So $\gcd (a^2,b^3) = 9^3$.

CASE 2: $a = 9k$ and $b = 9j$ and $3 \not \mid k$. Then $\gcd(a^2,b^3) = \gcd(9^2k^2, 9^3j^3)$. These have $9^2$ in common. $j$ and $k$ have nothing in common and no power of $3$ divides $k^2$.

So $\gcd(a^2,b^3) = 9^2$.

So the $\gcd(a^2,b^3) = 9^3$ if $27|a$ but $\gcd(a^2,b^3) = 9^2$ if $27 \not \mid a$.

Answer 4

If $\gcd(a, b) = 9$, that means that $a = 9m$, $b = 9n$, and $\gcd(m, n) = 1$.

Just in case you have any doubt on that last point, let's say $k = \gcd(m, n) > 1$. Then $k$ is a divisor of both $a$ and $b$, and in fact $\gcd(a, b) = 9k > 9$. But if $k = \gcd(m, n) = 1$, that preserves $\gcd(a, b) = 9$.

With that out of the way, we can tackle $\gcd(a^2, b^3)$. We see that $a^2 = 81m^2$ and $b^3 = 729n^3$. Since squaring and cubing add no new distinct prime factors, $\gcd(m^2, n^3) = 1$. This problem then boils down to $\gcd(81, 729) = 81$.

Try plugging in some specific values: $\gcd(18, 45) = 9$. Then $\gcd(18^2, 45^3) = \gcd(324, 91125) = 81$.