\documentclass[a4paper,12pt]{article}
\begin{document}
\parindent=0pt
QUESTION
Are the following true or false? Provide either a proof, or a
counterexample, as appropriate.
\begin{description}
\item[(i)]
If gcd$(a,p^2)=p$, then gcd$(a^2,p^2)=p^2$.
\item[(ii)]
If gcd$(a,p^2)=p$ and gcd$(b,p^2)=p^2$, then gcd$(ab,p^4)=p^3$.
\item[(iii)]
If gcd$(a,p^2)=p$ and gcd$(b,p^2)=p$ then gcd$(ab,p^4)=p^2$.
\item[(iv)]
If gcd$(a,p^2)=p$ then gcd$(a+p,p^2)=p$.
\end{description}
ANSWER
\begin{description}
\item[(i)]
TRUE: gcd$(a,p^2)=p\Rightarrow p^2|a^2$, so gcd$(a^2,p^2)=p^2$.
\item[(ii)]
FALSE: $gcd(a,p^2)=p$ shows that $p|a$ but $p\not|a$ so we know
the axact power of $p$ that divides $a$. But gcd$(b,p^c)=p^2$ only
shows that $p^2|b$. It is quite possible that $b$ is divisible by
a higher power of $p$ (e.g. gcd$(p^3,p^2)=p^2$) and it is by
exploring this possibility that we find a counterexample, e.g.
$a=2,\ b=8,\ p=2$ gives gcd$(a,p^2)$=gcd$(2,4)=2=p$ and
gcd$(b,p^2)$=gcd$8,4)=4=p^2$, but
gcd$(ab,p^4)$=gcd$(16,16)=16=p^4\neq p^3$.
\item[(iii)]
TRUE: As explaind in part (ii), we have $p|a$, but $p^2\not|a$, so
$a=pm$, where $p\not|m$. Similarly, $b=np$, where $p\not|n$. We
now have $ab=p^2mn$, and lemma 2.2 tells us that if $p|mn$ then
$p|m$ or $p|n$. Since we know that $p$ divided neither $m$ or $n$,
we may conclude that $p\not|mn$, so the highest power of $p$
dividing $ab$ is $p^2$. As the only divisors of $p^4$ are powers
of $p$, we may conclude that gcd$(ab,p^4)=p^2$.
\item[(iv)]
FALSE: The trick here is to take $a=p^2-p$, so that $a+p$ is
divisible by $p^2$. So to get a numerical counterexample, we may
choose, e.g.$p=3,\ a=6$ and get gcd$(a,p^2)$=gcd$(6,9)=3=p$, but
gcd$(a+p,p^2)$=gcd(9,9)=9.
\end{description}
\end{document}