(i) Prove that if $p$ and $q$ are distinct primes then $\gcd (p^n, q^n)=1$ for any natural number $n$.
(ii) Prove that if $p$ and $q$ are distinct primes then $\gcd (p^n, q^m) = 1$ for any natural numbers $m$ and $n$.
(i) Let $g = \gcd(p^n, q^n)$.
Since $p$ is prime, the only factors of $p^n$ are $$1, p, p^2, \ldots, p^n$$
Since $q$ is prime, the only factors of $q^n$ are $$1, q, q^2, \ldots, q^n$$
And so $g$ must be one of these factors from each list. That is, $g=p^j = q^k$ where $j,k$ is between 0 and $n$ inclusive.
If we assume $g>1$ for the purpose of contradiction, then $j,k$ is between 1 and $n$ inclusive. That is, we exclude $j=k=0$.
In this case, $g=p^j = q^k$ gives us $p \mid g$ and $g \mid q^k$. This gives us $p \mid q^k$. By Corollary 2.4, this means $p=q$. This is a contradiction since $p$ and $q$ are distinct.
Therefore $g>1$ was a false assumption, and so $\gcd (p^n, q^n)=1$.
(ii) Let $g = \gcd(p^n, q^m)$.
Since $p$ is prime, the only factors of $p^n$ are $$1, p, p^2, \ldots, p^n$$
Since $q$ is prime, the only factors of $q^n$ are $$1, q, q^2, \ldots, q^m$$
And so $g$ must be one of these factors from each list. That is, $g=p^j = q^k$ where $0 \le j \le n$ and $0 \le j \le m$.
If we assume $g>1$ for the purpose of contradiction, then $1 \le j \le n$ and $1 \le j \le m$.
Again, $g=p^j = q^k$ gives us $p \mid g$ and $g \mid q^k$. This gives us $p \mid q^k$. By Corollary 2.4, this means $p=q$. This is a contradiction since $p$ and $q$ are distinct.
Therefore $g>1$ was a false assumption, and so $\gcd (p^n, q^n)=1$.