MHB Calculate $(ab,p^4)$: Prime Number Exercise

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The exercise involves finding the greatest common divisor $(ab, p^4)$ given that $(a, p^2) = p$ and $(b, p^3) = p^2$. The discussion confirms that $a$ contains one factor of the prime $p$, while $b$ contains two factors of $p$, resulting in $ab$ having three factors of $p$. Consequently, the greatest common divisor is determined to be $\gcd(ab, p^4) = p^3$. The participants express understanding and appreciation for the clarification provided.
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Hey! :o
I am given the following exercise:
If $p$ is a prime and $(a,p^2)=p$,$(b,p^3)=p^2$,find $(ab,p^4)$.

$p|a \Rightarrow a=k \cdot p , k\in \mathbb{Z}$
$p^2|b \Rightarrow b=l \cdot p^2 , l \in \mathbb{Z}$

Let $(ab,p^4)=d>1$,then $d$ will have a prime divisor, $q$
$q|d , d|ab \Rightarrow q|ab \Rightarrow q|a \text{ or } q|b$
Also, $d|p^4 \Rightarrow q|p^4 \Rightarrow q=p$

Is it right so far?And how can I continue??
 
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evinda said:
Hey! :o
I am given the following exercise:
If $p$ is a prime and $(a,p^2)=p$,$(b,p^3)=p^2$,find $(ab,p^4)$.

$p|a \Rightarrow a=k \cdot p , k\in \mathbb{Z}$
$p^2|b \Rightarrow b=l \cdot p^2 , l \in \mathbb{Z}$

Let $(ab,p^4)=d>1$,then $d$ will have a prime divisor, $q$
$q|d , d|ab \Rightarrow q|ab \Rightarrow q|a \text{ or } q|b$
Also, $d|p^4 \Rightarrow q|p^4 \Rightarrow q=p$

Is it right so far?And how can I continue??

Hi! :)

It is right so far.

But to get further, I would try to identify how many factors $p$ are contained in $a$ exactly.
And also how many factors $p$ there are in $b$.

How many factors $p$ does that make in $ab$?
 
I like Serena said:
Hi! :)

It is right so far.

But to get further, I would try to identify how many factors $p$ are contained in $a$ exactly.
And also how many factors $p$ there are in $b$.

How many factors $p$ does that make in $ab$?

In $a$, there is $1$ factor $p$,in $b$ there are $2$ factors $p$.So,in $ab$ there are $3$ factors $p$..What does this mean?? :confused: Or am I wrong??
 
evinda said:
In $a$, there is $1$ factor $p$,in $b$ there are $2$ factors $p$.So,in $ab$ there are $3$ factors $p$..What does this mean?? :confused: Or am I wrong??

That is exactly right. :D

It means that $ab$ has exactly 3 factors of the prime $p$ and possibly some other prime factors.
The consequence is that $\gcd(ab, p^4) = p^3$.
 
I like Serena said:
That is exactly right. :D

It means that $ab$ has exactly 3 factors of the prime $p$ and possibly some other prime factors.
The consequence is that $\gcd(ab, p^4) = p^3$.

I understand..thank you very much! :cool:
 
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