## Task B-1.3.

Solve the equation in the set of prime numbers

$$
2 p^{3}-q^{2}=2(p+q)^{2}
$$
## Solution.

Since the right side of the equation is an even number, the left side must also be an even number. For the expression $2 p^{3}-q^{2}$ to be even, $q^{2}$ must be even, and thus $q$ must be an even number. The only even prime number is 2, so $q=2$. Then the original equation becomes

$$
2 p^{3}-4=2(p+2)^{2}
$$

which is equivalent to

$$
p^{3}-p^{2}-4 p-6=0
$$

or

$$
p\left(p^{2}-p-4\right)=6 \text{. }
$$

We conclude that $p$ must be a divisor of 6, and since $p$ is a prime number, it can only be $p=2$ or $p=3$.

It is easily verified that the above equation is satisfied only for $p=3$, so the ordered pair $(p, q)=(3,2)$ is the only solution to the given equation.