calculate the gcd of $a+b$ and $p^4$

Question

If $p$ is a prime and $(a,p^2)=p, (b,p^3)=p^2$,calculate $(a+b,p^4)$ $$$$ That's what I thought: $$$$ $$(a,p^2)=p \Rightarrow a=kp$$ $$(b,p^3)=p^2 \Rightarrow b=l p^2$$

So,$$(a+b,p)=(p+p^2)$$

Is it right or am I wrong?

Answer 1

The final equation is wrong. First, you need also $\,\color{#c00}{p\nmid k}.\,$ Therefore

$\qquad(a\!+\!b,\,p^4)\, = \,(kp\!+\!\ell p^2,\,p^4)\, =\, p\,(k\!+\!\ell p,\,p^3)\, =\, p,\,\ $ by $\,\ p\nmid k\!+\!\ell p\ $ (else $\,\color{#c00}{p\mid k})$

Answer 2

If you prefer, think in base $p$. Then $a$ ends with a single $0$ and $b$ ends with two $0$'s. So their sum ends with a single 0. That is, $(a+b,p^4)=p$.