0
$\begingroup$

Let $p$ be a prime. Suppose that $\gcd(a, b) = p$. Find $\gcd(a^2,b)$ for all integers a and b.

I was able to prove that $\gcd(a^2,b^2) = p$ but I don't know if it helps.

Any help is greatly appreciated.

$\endgroup$
1
  • 4
    $\begingroup$ Surely $\gcd (a^2,b^2)=p^2$. $\endgroup$
    – lulu
    May 30 '17 at 22:09
5
$\begingroup$

It depends:

If $p^2 \mid b$ then $\gcd (a^2,b)=p^2$

If $p^2 \nmid b$ then $\gcd(a^2,b)=p$

Given that $\gcd(a,b)=p$ and $p$ is a prime, clearly $a$ and $b$ share no prime factors except for a single $p$. So, squaring $a$ does not result in any further shared prime factors, except for possibly a second $p$, but that is only if $p^2 \mid b$.

$\endgroup$
2
$\begingroup$

Another way to restate Bram's answer is to say that $p\mid \gcd(a^2,b)\mid p^2$.

More generally, if $\gcd(a,b)=d$, then $d\mid \gcd(a^2,b)\mid d^2$.

The fact that $d\mid \gcd(a^2,b)$ is easy to see, so we'll just show that $\gcd(a^2,b)\mid d^2$.

Solve $ax+by=d$. Squaring this, we see that $a^2x^2+b(2axy+by^2)=d^2$. This means that $\gcd(a^2,b)\mid d^2$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.