a) Let us prove the following
Lemma. If for each integer $x$ the value of the polynomial $Q(x) = bx^2 + cx + d$ with integer coefficients is a square of an integer, then there exist integers $m$ and $n$ such that $b = m^2$, $c = 2mn$, $d = n^2$ and, in particular, $Q(x) = (mx + n)^2$.

Proof of the lemma. If $b = 0$, then for each integer $x$ the difference $Q(x+1) - Q(x) = c$ is constant. Whereas the differences between neighboring squares of integers infinitely grow, so if $c \ne 0$ then the non-constant polynomial $Q(x)$ takes the same values infinitely many times, which is impossible. If $b = c = 0$ then $d$ is a perfect square and the numbers $m = 0$ and $n = \sqrt{d}$ fit in.

Let $b \neq 0$, then, obviously, $b > 0$, since otherwise $Q(x)$ will be negative for sufficiently large $x$. If the discriminant $c^2 - 4bd$ is zero, then $Q(x) = b(x + \frac{c}{2b})^2$. Since $4b^2Q(x) = b(2bx + c)^2$ is a perfect square for every integer $x$, $b$ is a perfect square. Let $b = e^2$, then $Q(x) = (ex + \frac{c}{2e})^2$ is a perfect square for every integer value of $x$, which implies that $c$ is a multiple of $2e$ and $Q$ satisfies the lemma statement.

If the discriminant $c^2 - 4bd$ is not equal to zero, then among the prime divisors of the values $Q(x)$ in integer points choose a prime divisor $p$, greater than $|c^2 - 4bd|$. Let $Q(x)$ be a multiple of $p$. Consider $Q(x+p)$:
$$Q(x+p) = Q(x) + bp^2 + p(2bx + c). \quad (1)$$
Note that $(2bx + c)^2 = 4aQ(x) + c^2 - 4bd$ is not a multiple of $p$, so $2bx + c$ is not a multiple of $p$ and from (1) we obtain
$$v_p(Q(x+p)) = v_p(Q(x)) + 1,$$
which is impossible because $Q(x)$ and $Q(x+p)$ are perfect squares. Therefore, this case is impossible and the lemma is proved.

Let's move on to solving the problem itself. Denote the polynomial given in the condition by $P(x, y)$. We will show that it can be decomposed into a product of two polynomials of the first degree with rational coefficients. Let
$$P(x, y) = a_0x^2 + a_1y^2 + a_2xy + a_3x + a_4y + a_5.$$
Suppose $a_1 \neq 0$. Consider $P(x, y)$ as a square trinomial of the variable $y$:
$$P(y) = a_1y^2 + (a_2x + a_4)y + (a_0x^2 + a_3x + a_5).$$
Since this square trinomial has an integer root for any integer value $x$, its discriminant
$$D(x) = (a_2x + a_4)^2 - 4a_1(a_0x^2 + a_3x + a_5)$$
is a perfect square for all integer values of $x$. The degree of this polynomial is at most two, so we can apply the lemma to it. Hence $D(x) = (mx + n)^2$ for some integers $m$ and $n$. Then the polynomial $P(y)$ can be expanded into a product of two linear factors with rational coefficients $P(y) = a_1(y - y_1)(y - y_2)$, which can be written as
$$P(x, y) = \frac{1}{4a_1}(2a_1y - (a_2x + a_4) - (mx + n))(2a_1y - (a_2x + a_4) + (mx + n)).$$
A similar expansion can be written in the case $a_0 \neq 0$.

Suppose now that $a_0 = a_1 = 0$. Since this polynomial has degree 2, $a_2 \neq 0$ and
$$P(x, y) = a_2xy + a_3x + a_4y + a_5.$$
By the problem statement, for any integer value $x$ the number $y = \frac{a_3x+a_5}{a_2x+a_4}$ is an integer, because $P(x, y) = 0$ can be true only for it. Hence, for any integer value $x$ the number
$$a_2(a_3x + a_5) - a_3(a_2x + a_4) = a_2a_5 - a_3a_4$$
is a multiple of the denominator $a_2x + a_4$, which is possible only if $a_2a_5 = a_3a_4$. The number $t = \frac{a_3}{a_2}$ satisfies the equality $\frac{a_5}{a_4} = \frac{a_3}{a_2} = t$ (if $a_4 = 0$ then $a_5 = 0$, and the case $a_3 = 0$ is obviously impossible). Then $\frac{a_3x+a_5}{a_2x+a_4} = t$ for all integers $x$, which contradicts the problem statement for $b = t$. Therefore this case is impossible.

Thus, the polynomial $P(x, y)$ decomposes into a product of two polynomials of the first degree with rational coefficients, we write this equality in the form
$$M \cdot P(x, y) = (k_1x + l_1y + C_1)(k_2x + l_2y + C_2),$$
where all coefficients are integers. Note that the equations $k_1x + l_1y + C_1 = 0$ and $k_2x + l_2y + C_2 = 0$ are linear Diophantine equations and there are three alternatives for each of them: 1) the equation has no entire solutions; 2) the equation has the constant solution on some variable; 3) the general solution of the equation has the form $(x_0 + lt, y_0 - kt), t \in \mathbb{Z}$. Moreover, the second alternative is impossible according to the problem condition, and in the third possibility we can always assume that the coefficients in the pairs $(k_1, l_1)$ and $(k_2, l_2)$ are coprime.

If $|l_i| = 1$, then all solutions of $P(a, y) = 0$ belongs to the corresponding bracket, so the first possibility holds for the other bracket and, consequently, $|k_i|$ is also equal to one, which gives the required multiplier in the problem condition. The case $|k_i| = 1$ is considered similarly.

Consider the remaining case: all coefficients $k_i$ and $l_i$ are at least two in absolute value. Note that in the series of positive integers, each $|l_1|$-th number from the first bracket is a solution to the equation $P(x, b) = 0$, as well as each $|l_2|$-th from the second bracket. Hence $\frac{1}{|l_1|} + \frac{1}{|l_2|} = 1$ and similarly $\frac{1}{|k_1|} + \frac{1}{|k_2|} = 1$. Under our assumptions this is only possible for $|l_1| = |l_2| = |k_1| = |k_2| = 2$, which contradicts the fact that the coefficients must be coprime.

b) Consider the polynomial $P(x, y) = (3x - 2y)(3x - 4y + 1)(3x - 4y - 1)$. Each of its factors is a linear function, and the zeros of these factors are the pairs $(2k, 3k)$, $(4k + 1, 3k + 1)$ and $(4k - 1, 3k - 1)$, $k \in \mathbb{Z}$. It is easy to see that this polynomial satisfies the condition.