Problem 9.6. Find all pairs of natural prime numbers $p$, $q$, that satisfy the equation

$$
3 p^{4}+5 q^{4}+15=13 p^{2} q^{2}
$$
Solution. The numbers $p$ and $q$ cannot both be odd simultaneously. For verification, the congruence modulo 4 is used. The left side, $3 p^{4}+5 q^{4}+15 \equiv 3(\bmod 4)$, and the right side, $13 p^{2} q^{2} \equiv 1(\bmod 4)$.

The numbers $p$ and $q$ cannot both be even, i.e., each equal to 2. Indeed, in this case, the left side is an odd number, while the right side is an even number. It remains to examine the cases where one of the numbers $p, q$ is even, i.e., equal to 2.

Let $p=2$. The equation takes the form $48+5 q^{4}+15=52 q^{2}$, i.e., $5 q^{4}-52 q^{2}+63=0$. From the biquadratic equation obtained, we find $q^{2}=\frac{7}{5}$ and $q^{2}=9$. Only the value $q=3$ is a prime number. Thus, a solution to the given equation is $(p, q)=(2,3)$.

Now let $q=2$. The equation takes the form $3 p^{4}+80+15=52 p^{2}$, i.e., $3 p^{4}-52 p^{2}+105=0$. From the biquadratic equation obtained, we find $p^{2}=\frac{7}{3}$ and $p^{2}=15$. The equation has no solutions, not even among integers.

Thus, the given equation has a single solution: $(p, q)=(2,3)$.