Find all pairs of primes $p, q$ such that 3$p^2q+16pq^2$ equals to square of natural number

Question

Find all pairs of prime numbers $p, q$ such that 3$p^2q+16pq^2$ equals to square of natural number

My attempt: I've been trying to calculate equation through square root but now I'm stuck, please help

Answer 1

After factoring as $pq(3p+16q)=n^2$ We claim that p divides n and q divides n since both p and q divide $n^2$. Thus $n=k.pq$ for some kind belongs to natural numbers.

$3p+16q=k^2 pq 3/q+16/p=k^2$

from here we obtain p=2,q=3 and p and q are primes there are no more solutions.

Answer 2

Assume $3p^2q+16pq^2=k^2$

Since $q$ divides the LHS, it must divide $k^2$, and since $q$ is prime, it must be true that $q\mid k^2 \Rightarrow q^2\mid k^2$. Since $q^2\mid 16pq^2$, it follows that $q^2\mid 3p^2q \Rightarrow q\mid 3p^2$. But $q\not \mid p^2$, so $q\mid 3 \Rightarrow q=3$

Since $p$ divides the LHS, it must divide $k^2$, and since $p$ is prime, it must be true that $p\mid k^2 \Rightarrow p^2\mid k^2$. Since $p^2\mid 3p^2q$, it follows that $p^2\mid 16pq^2 \Rightarrow p\mid 16q^2$. But $p\not \mid q^2$, so $p\mid 16 \Rightarrow p=2$

This checks out, as $3\cdot 4\cdot 3+16\cdot 2\cdot 9=324=18^2$