## Task 13/88

Determine all pairs $(p ; q)$ of prime numbers $p$ and $q$ for which the following holds:

$$
3 p^{2}+6 p=2 q^{2}+7 q
$$
From two parabolas with different coefficients of the quadratic terms, the one with the larger coefficient grows more strongly. It follows immediately: If the given equation has solutions, then $q>p$. We write the equation in the form

$$
3 p(p+2)=q(2 q+7)
$$

From the uniqueness of the prime factorization, we now have three cases:

1. $q=3$; it follows that $p(p+2)=13$. Since 13 cannot be factored into two factors $\mathrm{p}$ and $p+2$, this case is ruled out.
2. $q=p$; it follows that $q=p=0$ or $q=p=1$. Since 0 and 1 are not prime numbers, this case is also ruled out.
3. $q=p+2$ (since $q>p$, $p+2$ cannot be a multiple of $q$); it follows that $p=11, q=13$ as the only solution pair.