Is the gcd of two numbers also the gcd of their squares?

Question

I'm working on a problem and I would like to use this fact, but I don't have anything to write on so I am having trouble figuring it out.

Is it always true that $\gcd(a, b) = \gcd(a^n, b^n)$?

I know it has to be true for prime numbers. I'm not sure about composite ones.

Obviously not. If $d$ divides both $a$ and $b$, $d^n$ divides both $a^n$ and $b^n$, hence $\gcd(a^n,b^n)\geq\gcd(a,b)^n$. You just need to show that $\leq $ holds, too. — Jack D'Aurizio, Oct 23 '15 at 20:38
probably confusing with the fact that, if $a$ and $b$ are coprime, then $a^n$ and $b^n$ are also coprime. — Bernard, Oct 23 '15 at 20:54
Note there's no contradiction: if $a$ and $b$ are coprime, their g.c.d. is equal to $1$, and $1^n=1$ for all $n$. — Bernard, Oct 23 '15 at 21:15
When trying to figure out things like this, it's very helpful to try some simple examples, like for example $ a = b = 2$. — littleO, Oct 23 '15 at 21:44

score 5 · Answer 1 · answered Oct 23 '15 at 21:38

5

Only if they're coprime to begin with, that is, $\gcd(a, b) = 1$. If the the two numbers share a common prime factor, $p$, then $a^2$ is divisible by $p^2$ and so is $b^2$.

An example to make it clearer: $\gcd(4, 14) = 2$ and $\gcd(16, 196) = 4$.

answered Oct 23 '15 at 21:38

David R.

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Is the gcd of two numbers also the gcd of their squares?

1 Answers1