Given that p is prime, $\gcd(a, p^2)=p$ and $\gcd(b, p^3)=p^2$ find $\gcd(a+b, p^4)$.
I'm really not sure how to approach the problem. My intuition from looking at it makes me think that the answer would simply be p. But I have no idea if that's correct or what the proper way to derive that answer is. Any help is greatly appreciated.
Since $p$ divides both $a$ and $b$, it divides $a+b$. You want to show that no higher power of $p$ divides $a+b$.
For contradiction, say $p^2$ divides $a+b$. Then, since $p^2$ also divides $b$, it followes that $p^2$ divides $(a+b)-b = a$. But...
Hint: Since $\gcd(a,p^2) = p$ and $\gcd(b,p^3)=p^2$, you know that $a = pm$ and $b = p^2n$ where $p\nmid m, n$. Then $a+b = p(m+pn)$, and $p\nmid (m+pn)$. Can you take it from there?
