Solve the equation $|x^2 -1|+ |x^2 - 4| = mx$ as a function of the parameter $m$. Which pairs $(x,m)$ of integers satisfy this equation?
To solve the equation \( |x^2 - 1| + |x^2 - 4| = mx \) as a function of the parameter \( m \), we need to analyze the behavior of the function \( f(x) = |x^2 - 1| + |x^2 - 4| \) and find the intersections with the line \( y = mx \).

### Step-by-Step Analysis:

1. **Define the function \( f(x) \):**
   \[
   f(x) = |x^2 - 1| + |x^2 - 4|
   \]
   Since \( f(x) \) is an even function, we only need to consider \( x \geq 0 \).

2. **Analyze the function in different intervals:**

   - **Case 1: \( 0 \leq x < 1 \)**
     \[
     f(x) = -(x^2 - 1) - (x^2 - 4) = -2x^2 + 5
     \]

   - **Case 2: \( 1 \leq x < 2 \)**
     \[
     f(x) = (x^2 - 1) - (x^2 - 4) = 3
     \]

   - **Case 3: \( 2 \leq x \)**
     \[
     f(x) = (x^2 - 1) + (x^2 - 4) = 2x^2 - 5
     \]

3. **Find the intersections with the line \( y = mx \):**

   - **For \( m < -3 \):**
     \[
     -2x^2 + 5 = mx \implies 2x^2 + mx - 5 = 0
     \]
     Solving the quadratic equation:
     \[
     x = \frac{-m \pm \sqrt{m^2 + 40}}{4}
     \]
     Since \( m < -3 \), we take the negative root:
     \[
     x = \frac{-m - \sqrt{m^2 + 40}}{4}
     \]

   - **For \( m = -3 \):**
     \[
     2x^2 - 5 = -3x \implies 2x^2 + 3x - 5 = 0
     \]
     Solving the quadratic equation:
     \[
     x = \frac{-3 \pm \sqrt{9 + 40}}{4} = \frac{-3 \pm 7}{4}
     \]
     Thus, \( x = 1 \) or \( x = -\frac{5}{2} \).

   - **For \( -3 < m < -\frac{3}{2} \):**
     \[
     2x^2 - 5 = mx \quad \text{or} \quad 3 = mx
     \]
     Solving \( 2x^2 - 5 = mx \):
     \[
     x = \frac{m - \sqrt{m^2 + 40}}{4}
     \]
     Solving \( 3 = mx \):
     \[
     x = \frac{3}{m}
     \]

   - **For \( m = -\frac{3}{2} \):**
     \[
     x = -2
     \]

   - **For \( -\frac{3}{2} < m < \frac{3}{2} \):**
     \[
     \text{No solution}
     \]

   - **For \( m = \frac{3}{2} \):**
     \[
     x = 2
     \]

   - **For \( \frac{3}{2} < m < 3 \):**
     \[
     -2x^2 + 5 = mx \quad \text{or} \quad 3 = mx
     \]
     Solving \( -2x^2 + 5 = mx \):
     \[
     x = \frac{3}{m} \quad \text{or} \quad x = \frac{m + \sqrt{m^2 + 40}}{4}
     \]

   - **For \( m = 3 \):**
     \[
     2x^2 - 5 = 3x \implies 2x^2 - 3x - 5 = 0
     \]
     Solving the quadratic equation:
     \[
     x = \frac{3 \pm \sqrt{9 + 40}}{4} = \frac{3 \pm 7}{4}
     \]
     Thus, \( x = 1 \) or \( x = \frac{5}{2} \).

   - **For \( m > 3 \):**
     \[
     -2x^2 + 5 = mx \implies 2x^2 + mx - 5 = 0
     \]
     Solving the quadratic equation:
     \[
     x = \frac{-m \pm \sqrt{m^2 + 40}}{4}
     \]
     Since \( m > 3 \), we take the positive root:
     \[
     x = \frac{-m + \sqrt{m^2 + 40}}{4}
     \]

4. **Identify integer solutions:**
   From the analysis, the pairs \((x, m)\) of integers that satisfy the given equation are:
   \[
   (x, m) = (-1, 3) \quad \text{and} \quad (1, 3)
   \]

The final answer is \( \boxed{ (x, m) = (-1, 3) \quad \text{and} \quad (1, 3) } \)