Problem 1503

Solution:
We write the equation in the form
$$a^{2} + 2 a p + p^{2} + b(3 a + 2 p) = 0$$
Hence
$$b = \frac{-(a + p)^{2}}{3 a + 2 p}$$
is an integer. This shows that $3 a + 2 p$ divides $(a + p)^{2}$ and hence also divides $(3 a + 3 p)^{2}$. But, we have
$$(3 a + 3 p)^{2} = (3 a + 2 p + p)^{2} = (3 a + 2 p)^{2} + 2 p(3 a + 2 p) + p^{2}$$
It follows that $3 a + 2 p$ divides $p^{2}$. Since $p$ is a prime, the only divisors of $p^{2}$ are $\pm 1, \pm p$ and $\pm p^{2}$. Since $p > 3$, we also have $p = 3 k + 1$ or $3 k + 2$.

Case 1: Suppose $p = 3 k + 1$. Obviously $3 a + 2 p = 1$ is not possible. In fact, we get $1 = 3 a + 2 p = 3 a + 2(3 k + 1) \Rightarrow 3 a + 6 k = -1$ which is impossible. On the other hand $3 a + 2 p = -1$ gives $3 a = -2 p - 1 = -6 k - 3 \Rightarrow a = -2 k - 1$ and $a + p = -2 k - 1 + 3 k + 1 = k$.
Thus $b = \frac{-(a + p)^{2}}{3 a + 2 p} = k^{2}$. Thus $(a, b) = \left(-2 k - 1, k^{2}\right)$ when $p = 3 k + 1$. Similarly, $3 a + 2 p = p \Rightarrow 3 a = -p$ which is not possible. Considering $3 a + 2 p = -p$, we get $3 a = -3 p$ or $a = -p \Rightarrow b = 0$. Hence $(a, b) = (-3 k - 1, 0)$ where $p = 3 k + 1$.
Let us consider $3 a + 2 p = p^{2}$. Hence $3 a = p^{2} - 2 p = p(p - 2)$ and neither $p$ nor $p - 2$ is divisible by $3$. If $3 a + 2 p = -p^{2}$, then $3 a = -p(p + 2) \Rightarrow a = -(3 k + 1)(k + 1)$.
Hence $a + p = (3 k + 1)(-k - 1 + 1) = -(3 k + 1) k$. This gives $b = k^{2}$. Again $(a, b) = (-(k + 1)(3 k + 1), k^{2})$ when $p = 3 k + 1$.

Case 2: Suppose $p = 3 k - 1$. If $3 a + 2 p = 1$, then $3 a = -6 k + 3$ or $a = -2 k + 1$. We also get
$$b = \frac{-(a + p)^{2}}{1} = \frac{-(-2 k + 1 + 3 k - 1)^{2}}{1} = -k^{2}$$
and we get the solution $(a, b) = \left(-2 k + 1, k^{2}\right)$. On the other hand $3 a + 2 p = -1$ does not have any integral solution for $a$. Similarly, there is no solution in the case $3 a + 2 p = p$. Taking $3 a + 2 p = -p$, we get $a = -p$ and hence $b = 0$. We get the solution $(a, b) = (-3 k + 1, 0)$. If $3 a + 2 p = p^{2}$, then $3 a = p(p - 2) = (3 k - 1)(3 k - 3)$ giving $a = (3 k - 1)(k - 1)$ and hence $a + p = (3 k - 1)(1 + k - 1) = k(3 k - 1)$. This gives $b = -k^{2}$ and hence $(a, b) = \left(3 k - 1, -k^{2}\right)$. Finally $3 a + 2 p = -p^{2}$ does not have any solution.