For all integers y, if $\gcd(9,y) = 3,$ then $\gcd(9,y^2) = 9$

Question

This question is based on the equation above, where for all integers $y$, if $\gcd(9,y) = 3$, then $\gcd(9,y^2) =9$.

I know this statement is true as if you try to put whatever that is a multiple of 3 as $y$ and if the value matched with the first condition where the $\gcd$ is $3$, it'll match the second condition. For example, if $y = 3,6,12,15,18,21,24...$ they will all work for both conditions.

What I'm trying to know now is how do I put into words to explain how they actually work in pair? More specifically, I am trying to understand the logic behind how they work for each other.

if $3$ divides $y$ then $9$ divides $y^2$ – J. W. Tanner Mar 26 '20 at 13:20 — J. W. Tanner, Mar 26 '20 at 13:20

score 0 · Accepted Answer · answered Mar 26 '20 at 13:23

0

Well, $\gcd(9,y^2)$ divides $9$, so it's $1, 3, $ or $9$.

If $\gcd(9,y)=3$ then $3$ divides $y$, so $9=3^2$ divides $y^2$.

Therefore, $\gcd(9,y^2)=9$.

answered Mar 26 '20 at 13:23

J. W. Tanner

60,406

For all integers y, if $\gcd(9,y) = 3,$ then $\gcd(9,y^2) = 9$

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