6. Find all pairs of prime numbers $p$ and $q$ such that $p^{2}+5 p q+4 q^{2}$ is a square of a natural number.
Answer: $(13,3),(7,5),(5,11)$

Solution. Let $p^{2}+5 p q+4 q^{2}=n^{2}$ for some natural number $n$. Then

$$
p q=n^{2}-(p+2 q)^{2}=(n-p-2 q)(n+p+2 q)
$$

The left side can be factored into the product of two numbers, one of which is an integer and the other a natural number, in four ways: $p q \cdot 1, p \cdot q, q \cdot p$ and $1 \cdot p q$. The number $n+p+2 q$ is greater than both $p$ and $q$, so the first three cases are impossible. Therefore, $n-p-2 q=1$ and $n+p+2 q=p q$. Thus, $2 p+4 q=p q-1$. Consequently, $(p-4)(q-2)=9$ and, therefore, $p-4= \pm 1, \pm 3, \pm 9$. Then $p=4 \pm 1,4 \pm 3$ or $4 \pm 9$. Among them, the prime numbers will only be $3,5,7$ and 13. The corresponding $q$ will be $-7,11,5$ and 3. Therefore, the first option does not work, but the others do.