`a ne 0, b ne 0`. The number of real number pair (a, b) which satisfy the equation `a^(4) + b^(4) = (a+b)^(4)` is:

Answer

To solve the equation \( a^4 + b^4 = (a + b)^4 \), we will follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ a^4 + b^4 = (a + b)^4 \]

### Step 2: Divide both sides by \( a^4 \)
Assuming \( a \neq 0 \) (as given), we can divide both sides by \( a^4 \):
\[ \frac{a^4 + b^4}{a^4} = \frac{(a + b)^4}{a^4} \]
This simplifies to:
\[ 1 + \left(\frac{b}{a}\right)^4 = \left(1 + \frac{b}{a}\right)^4 \]

### Step 3: Let \( x = \frac{b}{a} \)
Let \( x = \frac{b}{a} \). Then we can rewrite the equation as:
\[ 1 + x^4 = (1 + x)^4 \]

### Step 4: Expand the right-hand side using the binomial theorem
Using the binomial theorem, we expand \( (1 + x)^4 \):
\[ (1 + x)^4 = 1 + 4x + 6x^2 + 4x^3 + x^4 \]
Thus, we have:
\[ 1 + x^4 = 1 + 4x + 6x^2 + 4x^3 + x^4 \]

### Step 5: Simplify the equation
Subtract \( 1 + x^4 \) from both sides:
\[ 0 = 4x + 6x^2 + 4x^3 \]

### Step 6: Factor out common terms
We can factor out \( 2x \):
\[ 0 = 2x(2 + 3x + 2x^2) \]

### Step 7: Set each factor to zero
This gives us two cases:
1. \( 2x = 0 \) which implies \( x = 0 \) (or \( b = 0 \))
2. \( 2 + 3x + 2x^2 = 0 \)

### Step 8: Analyze the first case
The first case \( x = 0 \) implies \( b = 0 \), which contradicts the condition \( b \neq 0 \).

### Step 9: Analyze the second case
Now, we solve the quadratic equation:
\[ 2x^2 + 3x + 2 = 0 \]
Using the discriminant:
\[ D = b^2 - 4ac = 3^2 - 4 \cdot 2 \cdot 2 = 9 - 16 = -7 \]
Since the discriminant is negative, this quadratic equation has no real roots.

### Conclusion
Since both cases lead to no valid solutions, we conclude that there are no real number pairs \( (a, b) \) that satisfy the equation \( a^4 + b^4 = (a + b)^4 \).

Thus, the number of real number pairs \( (a, b) \) that satisfy the equation is:
\[ \boxed{0} \]