
NT4 Find all prime positive integers $p, q$ such that $2 p^{3}-q^{2}=2(p+q)^{2}$.


## Solution 1

The given equation can be rewritten as $2 p^{2}(p-1)=q(3 q+4 p)$.

Hence $p\left|3 q^{2}+4 p q \Rightarrow p\right| 3 q^{2} \Rightarrow p \mid 3 q$ (since $p$ is a prime number) $\Rightarrow p \mid 3$ or $p \mid q$. If $p \mid q$, then $p=q$. The equation becomes $2 p^{3}-9 p^{2}=0$ which has no prime solution. If $p \mid 3$, then $p=3$. The equation becomes $q^{2}+4 q-12=0 \Leftrightarrow(q-2)(q+6)=0$.

Since $q>0$, we get $q=2$, so we have the solution $(p, q)=(3,2)$.

