Suppose $p$ is a prime greater than $3$. Find all pairs of integers $(a,b)$ satisfying the equation $$a^2+3ab+2p(a+b)+p^2=0$$
A good way to start (probably) was to complete the square, $$p^2+2p(a+b)+a^2+2ab+b^2+ab-b^2=0$$ $$\implies (p+a+b)^2=b(b-a)$$ $$\implies ab=b^2-(p+a+b)^2$$ $$\implies ab=(p+a+2b)(-p-a)$$ But here I'm stuck, this last equality strongly suggests a solution in negative integers, but I am unable to find it..
Aside from this, I have also managed to obtain that $$b(2p+3a)<0$$ I don't know if it helps though..
HINT
$$a^2+(2p+3b)a+p^2+2pb=0$$
Since $a$ is a integer, $\Delta$(the polynomial discriminant) is a square.
This implies $$(2p+3b)^2-4(p^2+2pb)=x^2$$For some integer $x$.
Thus $$4pb+9b^2=x^2 \Leftrightarrow 81b^2+36bp=9x^2 \Leftrightarrow (9b+2p)^2-4p^2=9x^2$$
Now, $$(9b+2p)^2-9x^2=4p^2$$
However, soltuions to $4p^2=t^2-u^2=(t+u)(t-u)$ are $(|t|,|u|)=( 2p, 0)$ and $( p^2+1, p^2-1)$