$\gcd(a+b, p^4)$ if $\gcd(a,p^2) = p$ and $\gcd(b,p^3)=p^2$

Question

I'm having some trouble solving the following exercice:

Let $a,b \in \mathbb Z$ and $p$ be a prime number. If $\gcd(a,p^2) = p$ and $\gcd(b,p^3)=p^2$ determine:

$\gcd(ab,p^4)$

$\gcd(a+b, p^4)$

I was able to determine the first one and got $\gcd(ab,p^4) = p^3$ (The part I'm having trouble with is the second one, but I would appreciate If someone could also confirm this first answer).

I'm not being able to determine the value of $\gcd(a+b, p^4)$. How can I do this?

score 2 · Accepted Answer · answered Dec 29 '20 at 16:25

2

Hint: From the given information, we can be sure that there are integers $x$ and $y$ for which $a = px$, $b = p^2 y$, and $\gcd(x,p) = \gcd(y,p) = 1$ (i.e. $x,y$ are not divisible by $p$). In terms of $x$ and $y$, we have $$ ab = p^3 xy, \quad a + b = px + p^2y = p(x + py). $$ How do we know that $x + py$ is not divisible by $p$?

answered Dec 29 '20 at 16:25

Ben Grossmann

225,327

Because $\gcd(p, x+py) = \gcd(p,x) = 1$? therefore $p \not | x+yp$. Is this correct? – Eduardo Magalhães Dec 29 '20 at 16:32
@EduardoMagalhães Yes, that is correct – Ben Grossmann Dec 29 '20 at 18:52
Please search first to avoid adding more dupe answers to FAQs – Bill Dubuque Dec 30 '20 at 11:41

$\gcd(a+b, p^4)$ if $\gcd(a,p^2) = p$ and $\gcd(b,p^3)=p^2$

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