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

[evinda]

Hey!
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??

 

[I like Serena]

Hey!
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 $ab$?

 

[evinda]

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?? Or am I wrong??

 

[I like Serena]

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?? 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 $\gcd(ab, p^4) = p^3$.

 

[evinda]

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!