If $\sqrt[5]{n}Let $n$ is a non zero perfect square, and $p$ is a smallest prime factor of $n$.
If $\sqrt[5]{n}<p$, then prove that either $\sqrt{n}$ is prime or $\sqrt{\frac{n}{p^2}}$ is prime.
Attempt: Suppose that $\sqrt{\frac{n}{p^2}}$ is composite. Then it has a least divisor $q$. Then $q\leq \sqrt{\frac{n}{p^2}}$$\Rightarrow$ $q^2\leq \frac{n}{p^2}$. Given $p^5>n$. So $q^2<p^3$.
I didn't get any contradiction. Kindly help me
 A: With $p^5 \gt n$ and $p$ being a smallest prime factor of $n$, this means $n$ has at most $4$ prime factors. Since $n$ is a non-zero perfect square, the number of prime factors of $n$ is a positive even integer, so it must be either $2$ or $4$.
If $n$ has just $2$ prime factors, then $n = p^2$, so $\sqrt{n} = p$ is prime, while $\sqrt{\frac{n}{p^2}} = 1$ is not prime. If, instead, $n$ has $4$ prime factors, then $n = p^{2}q^{2}$ for some prime $q \ge p$ (i.e., it may be that $q = p$), which means $\sqrt{n} = pq$ is composite and $\sqrt{\frac{n}{p^2}} = q$ is prime.
A: Even if $\sqrt{\frac{n}{p^2}}$ were prime, it would still have a least divisor $q$, so you never used your assumption $\sqrt{\frac{n}{p^2}}$ is composite.
A proof along the same vein would go:
Suppose $\sqrt{\frac{n}{p^2}}$ is composite.  Then there are two primes (perhaps equal) $q_1, q_2$ such that $q_1q_2$ divides $\sqrt{\frac{n}{p^2}}$.
Then $q_1^2q_2^2$ divides $\frac{n}{p^2}$ and therefore divides $n$.  Since $p$ is the smallest prime divisor of $n$, we see
$$p^2p^2 \leq q_1^2q_2^2 \leq \frac{n}{p^2} < \frac{p^5}{p^2} = p^3 \\ \implies p^4 < p^3$$
which is your contradiction.
You can conclude either $\sqrt{\frac{n}{p^2}}$ is prime or it's equal to $1$, in which case $\sqrt{n} = p$ is prime and we're finished.
