Solution of polynomial $^2$

Assume $\left[Q(x)\right]^2 = 1$ has four positive integral solutions. Then $\begin{aligned}x^2 + ax + b - 1 &= 0 \\ x^2 + ax + b + 1 &= 0\end{aligned}$ have two positive integral solutions each.

Let $s \leq s_1$ denote the two integral solutions of the first equation, so that $s+s_1 = -a$ and $ss_1 = b-1$ . Since $1 \leq s\leq s_1$ by assumption, we see that $s(-a-s)=b-1, \qquad s \leq -\frac{a}{2}.$
Similarly, if $t \leq t_1$ denote the two integral solutions of the second equation, then $t(-a-t)=b+1, \qquad t \leq -\frac{a}{2}.$

Since $x\mapsto x(-a-x)$ is increasing on $x \leq -\frac{a}{2}$ and $s(-a-s) < t(-a-t)$ , it follows that $s < t$ . Also, we see that $\begin{aligned} 2 &= (b+1)-(b-1) \\ &= t(-a-t)-s(-a-s) \\ &= -a(t-s) - (t^2-s^2) \\ &= (t-s)(-a - t - s) \end{aligned}$ so that $\{t-s, -a-t-s\} = \{1,2\}$ , or $\begin{aligned} 3 &= 1+2 = (t-s)+(-a-t-s) = -a-2s \\ &\implies a = -2s-3 \\ &\implies b = 1+s(-a-s) = 1+s(2s+3-s) = s^2 + 3s + 1 \end{aligned}$

At this point, all we have shown is that if $a,b$ are such that $\left[Q(x)\right]^2 = 1$ has four positive integer solutions, then $(a,b) \in \{(-2s-3, s^2+3s+1) : s\in \mathbb{Z}, 1 \leq s\}$ However, for such values we can directly factor $\begin{aligned} 0 &= \left[Q(x)\right]^2 - 1 \\ &= (x^2 - (2s+3)x + (s^2+3s+1) - 1)(x^2 - (2s+3)x + (s^2+3s+1) + 1) \\ &= (x-s)(x-(s+3))(x-(s+1))(x-(s+2)) \end{aligned}$ so this condition on the coefficients $a,b$ is actually both necessary and sufficient.

The only requirement we haven't used is $|b| \leq 365$ . Therefore, we solve $1\leq s, \qquad |s^2+3s+1| \leq 365$ for integral $s$ , which gives $1\leq s \leq 17.$

Therefore, there are $\boxed{17}$ such pairs $(a,b)$ satisfying the assumptions in the problem.