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:
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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