How can I prove that the nth-root of $PQ$, where $P$ and $Q$ are prime, is irrational? Here is my proof for the case n=2.
Proof: Assume, to the contrary, that $\sqrt{pq}$ is rational. Then $\sqrt{pq}$ $=\dfrac{x}{y}$ for two integers $x$ and $y$ and we further assume that $gcd(x,y)=1$.
Now, $pq=\frac{x^2}{y^2}$ and so $pqy^2=x^2$.
Observe that $p∣x^2$ and by Euclid's Lemma, $p∣x$. Thus $x=pk$ for some integer $k$ and so $pqy^2=x^2=(pk)^2$ and so $qy^2=pk^2$. Hence, $p∣qy^2$. This implies that with $p|y^2$ or $p|q$. Since $p$,$q$ are distinct primes, $p≠q$ and so $p∤q$, which means $p∣y^2$. Thus, $p∣y$. This contradicts our assumption that $gcd(x,y)=1$.
How can I extend this to the nth-root?
 A: I'll write out a detail or two of the solution I suggested in the comments.
You should be able to take your proof for $n=2$, and rewrite it line-by-line using the same steps, to extend to the general case $n \ge 2$.
So, as you said in the comments, starting from the assumption $\displaystyle \sqrt[n]{pq} = \frac{x}{y}$ for two integers $x$ and $y$ such that $gcd(x,y)=1$, you deduce in the same steps that $q y^n = p^{n-1} k^n$.
The next step is new: knowing that $n \ge 2$, it follows that $n-1 \ge 1$, and so
$$p \mid p^{n-1} \mid p^{n-1} k^n = q y^n
$$
From $p \mid q y^n$, it follows that $p \mid q$ or $p \mid y^n$, and now you continue on exactly as before.
A: If $pq=m^n$ consider the prime factorization on each side.
On the left,
each prime has exponent $1$.
On the right,
if $r$ is a prime dividing $m$,
then $r|pq$ so
$r|p$ or $r | q$.
If $r|p$ then
$r \not\mid q$.
Since $r$ and $p$ are both primes,
$r=p$.
Therefore,
if $m_1=m/r$,
$rq
=m^n=(rm_1)^n
=r(r^{n-1})m_1^n
$
so
$q=(r^{n-1})m_1^n$
which cannot be by unique factorization.
Similarly if $r|q$.
Even more simply.
If $pq=m^n$ consider the prime factorization on each side.
On the left side
all the exponents are $1$
while on the right side
all the exponents are at least $n$.
Note that this argument
works for any number
of distinct primes on the left
and any $n \ge 2$.
