1. Find all pairs $(a, b)$ of integers that satisfy the equation

$$
a^{2}+7 a b+6 b^{2}+5 a+4 b+3=0 .
$$
SOLUTION. We solve the equation as a quadratic with the unknown $a$ and parameter $b$. Its discriminant is

$$
D=(7 b+5)^{2}-4\left(6 b^{2}+4 b+3\right)=25 b^{2}+54 b+13
$$

and the roots are

$$
a_{1,2}=\frac{-7 b-5 \pm \sqrt{D}}{2}
$$

If $a$ and $b$ are integers, then $\sqrt{D}= \pm(2 a+7 b+5)$ must also be an integer. Therefore, we can write

$$
D=25 b^{2}+54 b+13=c^{2},
$$

where $c$ is a non-negative integer. The equation

$$
25 b^{2}+54 b+13-c^{2}=0
$$

is again solved as a quadratic. Its roots are

$$
b_{1,2}=\frac{-27 \pm \sqrt{27^{2}-25 \cdot 13+25 c^{2}}}{25} .
$$

If $b$ and $c$ are integers, then $\sqrt{404+25 c^{2}}$ must be the square of some non-negative integer $d$. For non-negative integers $c, d$, we have $d^{2}-25 c^{2}=404$ or

$$
(d+5 c)(d-5 c)=404
$$

The difference $(d+5 c)-(d-5 c)=10 c$ is even, so the numbers $d+5 c$ and $d-5 c$ have the same parity. Furthermore, for non-negative $c$, $d+5 c \geqq d-5 c$ and $d+5 c \geqq 0$, so the only suitable factorization of 404 into two integers is

$$
d+5 c=202, \quad d-5 c=2 .
$$

From this, $d=102, c=20$. From the roots

$$
b_{1,2}=\frac{-27 \pm d}{25}
$$

only $b=3$ is an integer. Then

$$
a_{1,2}=\frac{-7 b-5 \pm c}{2}
$$

so $a_{1}=-3$ and $a_{2}=-23$.

The given equation is satisfied by two pairs of numbers $(a, b)$, namely $(-3,3)$ and $(-23,3)$.