Let $f(x)=|x^2-1|+|x^2-4|$, $f(x)$ is an even function, thus we are to cinsider the graph of $y=f(x)$ for $x\geq 0$. We are to examine of the coordinate of $x$ of $C: y=f(x)$ and the line $l:y=mx$. Case 1 : $0\leq x<1$ $f(x)=-(x^2-1)-(x^2-4)\therefore f(x)=-2x^2+5$. Case 2 : $1\leq x<2$ $f(x)=(x^2-1)-(x^2-4)\therefore f(x)=3$. Case 3 : $2\leq x$ $f(x)=(x^2-1)+(x^2-5)\therefore f(x)=2x^2-5$. Therefore we get the graph of $y=f(x)$ as the attached figure. (I will post it later.) Let $A(-1,\ 3),\ A'(1,\ 3),\ B(-2,\ 3),\ B'(2,\ 3)$, we have : $A\in{l}\Longleftrightarrow m=-3$ $B\in{l}\Longleftrightarrow m=-\frac 32$ $B'\in{l}\Longleftrightarrow m=\frac 32$ $A'\in{l}\Longleftrightarrow m=3$ Therefore, the answer is as follows : $1^\circ\ m<-3$ Solving $-2x^2+5=mx\Longleftrightarrow 2x^2+mx-5=0$, by the graph, gives ${x=\frac{-m\pm \sqrt{m^2+40}}{4}}$ by the graph, $x=\boxed{\frac{-m-\sqrt{m^2+40}}{4}}$ $2^\circ\ m=-3$ Solving $2x^2-5=-3x$ gives $x=\boxed{-\frac 52,\ -1}$ $3^\circ\ -3<m<-\frac 32$ Solving $2x^2-5=mx$ or $3=mx$, by the graph, gives $x=\boxed{\frac{m-\sqrt{m^2+40}}{4},\ \frac{3}{m}}$ $4^\circ\ m=-\frac 32$ By the graph, we have $x=\boxed{-2}$ $5^\circ\ -\frac 32<m<\frac 32$ $\boxed{\text{No Solution}}$ $6^\circ\ m=\frac 32$ By the graph, we have $x=\boxed{2}$ $7^\circ\ \frac 32<m<3$ Solving $-2x^2+5=mx$ or $3=mx$, by the graph, gives $x=\boxed{\frac{3}{m},\ \frac{m+\sqrt{m^2+40}}{4}}$ $8^\circ\ m=3$ Solving $2x^2-5=3x$ gives $x=\boxed{1,\ \frac 52}$ $9^\circ\ m>3$ Solving $-2x^2+5=mx\Longleftrightarrow 2x^2+mx-5=0$, by the graph gives ${x=\frac{-m\pm \sqrt{m^2+40}}{4}}$ by the graph, $x=\boxed{\frac{-m+\sqrt{m^2+40}}{4}}$ Consequently, from the result, the pairs $(x,\ m)$ of integers which satisfy the given equation are $(x,\ m)=\boxed{(-1,3),\ (1,\ 3)}$

$(5,9),\, (-5,9)$ are also solutions! And it is much easier to write $m=\frac{|x^2-1| + |x^2-4|}{x}$ and to proceed: Case 1. $ 0\leq x<1 $, $m= -2x+\frac{5}{x}\, \Rightarrow x | 5$, ... Case 2. $ 1\leq x<2 $, $m=\frac{3}{x}$, ... Case 3. $ 2\leq x $, $ m=2x - \frac{5}{x}$, ...