I have the following question from INMO 2004, Find all pairs of integers (a,b) satisfying the equation, where $p$ is prime
$$a^2 + 3ab + 2p(a+b) + p^2 = 0$$
for which I proceed this way:
$$(a+b+p)^2 = b^2 - ab$$
which means $b^2 - ab$ is a perfect square. Let $gcd(a,b) = t$ where $a = st$ and $b = rt$ where $gcd(r,s) = 1$. which means $t^2r(r-s)$ is a perfect square however $gcd(r,r-s) = 1$ which means both $r$ and $r-s$ are perfect squares. Let $r = \alpha^2$and $s=\alpha^2 - \beta^2$ which reduces the above equation to $$(a+b+p)^2 = \alpha^2\beta^2t^2$$
which implies either $$a + b + p = \alpha\beta t$$or $$a+b+p = -\alpha\beta t$$
again substituting a and b as above in terms of $\alpha,\beta,t$ I get $$p = t(\beta - \alpha)(2\alpha + \beta)$$ in first case and $$p = t(\alpha + \beta)(-2\alpha + \beta)$$ in the second case. Since $p$ is a prime, we have two terms of the above expression to be one. Upon solving I can get $3$ solutions, two of which are obtained specifically only when $p$ is congruent to $1$ (mod $3)$. Similarly, upon solving the second case above, I again end up with same set of solutions for $(a,b)$ again specifically for $p$ congruent to $1$ (mod $3$).
However my book says solutions exist even for $p$ congruent to $-1$ mod $3$. However They aren't coming up in my solution, why is it so? Have I missed any case?