What are all pairs of integers $(r, p)$ for which $r^{2}-r(p+6)+p^{2}+5 p+6=0$ ?
We consider $r^{2}-r(p+6)+p^{2}+5 p+6=0$ to be a quadratic equation in $r$ with two coefficients that depend on the variable $p$.

For this quadratic equation to have real numbers $r$ that are solutions, its discriminant, $\Delta$, must be greater than or equal to 0 . A non-negative discriminant does not guarantee integer solutions, but may help us narrow the search.

By definition,

$$
\begin{aligned}
\Delta & =(-(p+6))^{2}-4 \cdot 1 \cdot\left(p^{2}+5 p+6\right) \\
& =p^{2}+12 p+36-4 p^{2}-20 p-24 \\
& =-3 p^{2}-8 p+12
\end{aligned}
$$

Thus, we would like to find all integer values of $p$ for which $-3 p^{2}-8 p+12 \geq 0$. The set of integers $p$ that satisfy this inequality are the only possible values of $p$ which could be part of a solution pair $(r, p)$ of integers. We can visualize the left side of this inequality as a parabola opening downwards, so there will be a finite range of values of $p$ for which this is true.

By the quadratic formula, the solutions to the equation $-3 p^{2}-8 p+12=0$ are

$$
p=\frac{8 \pm \sqrt{8^{2}-4(-3)(12)}}{2(-3)}=\frac{8 \pm \sqrt{208}}{-6} \approx 1.07,-3.74
$$

Since the roots of the equation $-3 p^{2}-8 p+12=0$ are approximately 1.07 and -3.74 , then the integers $p$ for which $-3 p^{2}-8 p+12 \geq 0$ are $p=-3,-2,-1,0,1$. (These values of $p$ are the only integers between the real solutions 1.07 and -3.74 .)

It is these values of $p$ for which there are possibly integer values of $r$ that work.

We try them one by one:

- When $p=1$, the original equation becomes $r^{2}-7 r+12=0$, which gives $(r-3)(r-4)=0$, and so $r=3$ or $r=4$.
- When $p=0$, the original equation becomes $r^{2}-6 r+6=0$. Using the quadratic formula, we can check that this equation does not have integer solutions.
- When $p=-1$, the original equation becomes $r^{2}-5 r+2=0$. Using the quadratic formula, we can check that this equation does not have integer solutions.
- When $p=-2$, the original equation becomes $r^{2}-4 r=0$, which factors as $r(r-4)=0$, and so $r=0$ or $r=4$.
- When $p=-3$, the original equation becomes $r^{2}-3 r=0$, which factors as $r(r-3)=0$, and so $r=0$ or $r=3$.

Therefore, the pairs of integers that solve the equation are

$$
(r, p)=(3,1),(4,1),(0,-2),(4,-2),(0,-3),(3,-3)
$$

ANSWER: $(3,1),(4,1),(0,-2),(4,-2),(0,-3),(3,-3)$