Conjecture: $ \gcd(\operatorname{rad}(a+b) ,ab)= \operatorname{rad}(\gcd(a,b))$

Question

I have discovered some exercise type conjectures which I can't prove and this is one of them:

Given positive integers $a,b$, then $$ \gcd(\operatorname{rad}(a+b) ,ab)= \operatorname{rad}(\gcd(a,b))$$

Can this be proved or disproved?

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From time to time, when testing my growing math packages BigZ and Forthmath, I recognize some patterns which I can't prove or disprove (or even have the ambition to). I post them here with the hope that it will not annoy too much. I hope you can bear with it.

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Answer 1

First note that the left- and right-hand sides of the conjectured equality are both square-free, since $\text{rad}(n)$ is always square-free. From the definition of $GCD$, it then suffices to check that a prime divides the left iff it divides the right.

Now since $p \mid \text{rad}(n) \iff p \mid n$, we see that $p \mid \text{rad}(a+b), ab \iff p \mid a+b, ab \iff p \mid a,b \iff p \mid \text{rad}(a), \text{rad}(b)$ and we are done.

Answer 2

https://en.wikipedia.org/wiki/P-adic_order

take $\nu_p(a) \geq \nu_p(b)$

(I) If $p$ divides both, then $\nu_p \operatorname{rad} (a+b) = 1,$ next $\nu_p \gcd( \operatorname{rad} (a+b), ab) = 1,$ finally $\nu_p(\operatorname{rad}\gcd(a,b) ) = 1$

(II) If $p$ divides $a$ but not $b,$ then $\nu_p \operatorname{rad} (a+b) = 0,$ next $\nu_p \gcd( \operatorname{rad} (a+b), ab) = 0,$ finally $\nu_p(\operatorname{rad}\gcd(a,b) ) = 0$

(III) if $p$ divides neither, both sides give zero again, despite the possibility that $p$ does divide $a+b.$

Answer 3

$$\gcd(\operatorname{rad}(a+b),ab)=\operatorname{rad}(\gcd(a,b))\tag{1}$$

Case 1: If $\gcd(a,b)=1$ then if $p\mid a+b$ and $p\mid ab$, then $p\mid a$ or $b$. If $p\mid a$ then $p\mid a+b$ implies $p\mid(a+b)-a=b$, a contradiction since $\gcd(a,b)=1$. In this case both sides of $(1)$ equal $1$ and the claim is true.

Case 2: If, for some prime $p$, $p\mid a$ and $p\mid b$, then $p\parallel \operatorname{rad}(\gcd(a,b))$. Let $P$ be the square-free product of all such prime divisors common to $a$ and $b$. Hence $P=\operatorname{rad}(\gcd(a,b))$, and we also have $P^2\mid ab$. Let $a=a_1P$ and $b=b_1P$, where $a_1$, $b_1\ge1$, so $a+b=a_1P+b_1P=P(a_1+b_1)$.

Now $\operatorname{rad}(a+b)=\operatorname{rad}(P(a_1+b_1))$. Next, $a_1+b_1$ will have some prime $q\mid a_1+b_1$, where $q\nmid P$. But does $q\mid a$ or $q\mid b$, it cannot divide both or it should be a factor of $P$. For this let $a+b=Pqt$, for some integer $t\ge1$, and $q\mid a$. Then $q\mid (a+b)-a=b$, a contradiction. So $P=\gcd(\operatorname{rad}(a+b),ab)$ and the claim is proved.