If $A$ is a positive integer and $\gcd(pq,A)=1$ holds (where $p$ and $q$ are primes), does it follow that $\gcd(p,A)=1$ and $\gcd(q,A)=1$?

Question

My question here is pretty basic, so I apologize in advance if there is a simple answer.

INITIAL QUESTION

If $A$ is a positive integer and $\gcd(pq,A)=1$ holds (where $p$ and $q$ are primes), does it follow that $\gcd(p,A)=1$ and $\gcd(q,A)=1$?

MY ATTEMPT

Let $A$ be a positive integer, and suppose that $\gcd(pq,A)=1$ holds (where $p$ and $q$ are primes).

Without loss of generality, suppose that $\gcd(p,A) > 1$. Since $p$ is prime, then we consider two cases:

Case 1: $A < p$

Since $p$ is prime, $A < p$ would imply that $\gcd(p,A) = 1$, contradicting our assumption.

Case 2: $A \geq p$

Since $p$ is prime, and since $\gcd(p,A) > 1$, then $A \geq p$ implies that $\gcd(p,A)=p$, which further means that $p \mid A$. This implies that we can write $A = pa$, for some positive integer $a$. Thus, we obtain $$\gcd(pq,A)=\gcd(pq,pa)=p\cdot\gcd(q,a) \geq p \geq 2,$$ which contradicts $\gcd(pq,A)=1$.

QED

FINAL QUESTION

Does the proof above generalize to a valid argument for the following proposition?

If $B$ is a positive integer and $$C = \prod_{i=1}^{\omega(C)}{p_i}$$ is squarefree such that $\gcd(B,C)=1$, does it follow that $\gcd(B,p_i)=1$ for each $i$, $1 \leq i \leq \omega(C)$?

an alternative definition of coprime is a Bezout identity. You begin with $pqx + Ay=1.$ But then $p(qx) + Ay=1$ and $q(px) + Ay=1.$ — Will Jagy, Jul 31 '22 at 13:14
@WillJagy: Thank you for your comment. So are you saying that Bezout's Identity directly proves $$\gcd(pq,A) = 1 \implies \bigl(\gcd(p,A) = 1 \land \gcd(q,A) = 1\bigr)?$$ And that it seems the hypothesis that $p$ and $q$ are primes is not even needed in order for the proof to work? — Jose Arnaldo Bebita Dris, Jul 31 '22 at 13:18
yes. you might look into why this is an equivalent definition; in one direction it is the Extended Euclidean Algorithm — Will Jagy, Jul 31 '22 at 13:22
For any $a,b,c$, a number that divides $b$ will also divide $bc$. Therefore $\gcd(a,b)$ necessarily divides $\gcd(a,bc)$ and the result follows. — eyeballfrog, Jul 31 '22 at 13:24

score 0 · Accepted Answer · edited Aug 13 '22 at 19:25

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Just note that any common divisor of $p$ and $A$ is also a common divisor of $pq$ and $A$. So $\gcd(pq,A)\ge \gcd(p,A) \ge 1$. But you're given that $\gcd(pq,A)=1$.

By the way $p$ and $q$ don't have to be primes for the result to hold.

edited Aug 13 '22 at 19:25

insipidintegrator

6,642

answered Jul 31 '22 at 13:25

paw88789

40,402

2

Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Jul 31 '22 at 14:32

If $A$ is a positive integer and $\gcd(pq,A)=1$ holds (where $p$ and $q$ are primes), does it follow that $\gcd(p,A)=1$ and $\gcd(q,A)=1$?

1 Answers1