How many ordered pair of integers ( a,b) satisfy all the following inequalities `a^(2) +b^(2) lt 16 , a^(2) +b^(2) lt 8 a,a^(2) +b^(2) lt 8b ` ?

Answer

To solve the problem of finding how many ordered pairs of integers (a, b) satisfy the inequalities \( a^2 + b^2 < 16 \), \( a^2 + b^2 < 8 \), and \( a^2 + b^2 < 8b \), we will analyze each inequality step by step.

### Step 1: Analyze \( a^2 + b^2 < 16 \)

This inequality describes a circle with radius 4 centered at the origin (0, 0). The integer points (a, b) that lie inside this circle must satisfy \( a^2 + b^2 < 16 \).

- The possible integer values for \( a \) and \( b \) can be found by checking values from -4 to 4 (since \( 4^2 = 16 \)).
- We will check each integer value of \( a \) and find the corresponding integer values of \( b \) that satisfy the inequality.

1. **For \( a = 0 \)**:
- \( b^2 < 16 \) → \( b \) can be -3, -2, -1, 0, 1, 2, 3 (7 values).

2. **For \( a = 1 \)**:
- \( 1^2 + b^2 < 16 \) → \( b^2 < 15 \) → \( b \) can be -3, -2, -1, 0, 1, 2, 3 (7 values).

3. **For \( a = 2 \)**:
- \( 2^2 + b^2 < 16 \) → \( b^2 < 12 \) → \( b \) can be -3, -2, -1, 0, 1, 2 (6 values).

4. **For \( a = 3 \)**:
- \( 3^2 + b^2 < 16 \) → \( b^2 < 7 \) → \( b \) can be -2, -1, 0, 1, 2 (5 values).

5. **For \( a = 4 \)**:
- \( 4^2 + b^2 < 16 \) → \( b^2 < 0 \) → No valid values for \( b \).

Now, we can summarize the counts:
- For \( a = 0 \): 7 values
- For \( a = 1 \): 7 values
- For \( a = 2 \): 6 values
- For \( a = 3 \): 5 values
- For \( a = -1 \): 7 values (by symmetry)
- For \( a = -2 \): 6 values
- For \( a = -3 \): 5 values
- For \( a = -4 \): 0 values

Calculating the total:
- Total pairs from \( a = 0 \) to \( a = 3 \): \( 7 + 7 + 6 + 5 = 25 \)
- Total pairs from \( a = -1 \) to \( a = -3 \): \( 7 + 6 + 5 = 18 \)

Thus, for the first inequality, we have \( 25 + 18 = 43 \) pairs.

### Step 2: Analyze \( a^2 + b^2 < 8 \)

This inequality describes a circle with radius \( \sqrt{8} \) (approximately 2.83) centered at the origin.

1. **For \( a = 0 \)**:
- \( b^2 < 8 \) → \( b \) can be -2, -1, 0, 1, 2 (5 values).

2. **For \( a = 1 \)**:
- \( 1^2 + b^2 < 8 \) → \( b^2 < 7 \) → \( b \) can be -2, -1, 0, 1, 2 (5 values).

3. **For \( a = 2 \)**:
- \( 2^2 + b^2 < 8 \) → \( b^2 < 4 \) → \( b \) can be -1, 0, 1 (3 values).

4. **For \( a = -1 \)** and \( a = -2 \): Same counts as \( a = 1 \) and \( a = 2 \).

Calculating the total:
- For \( a = 0 \): 5 values
- For \( a = 1 \): 5 values
- For \( a = 2 \): 3 values
- For \( a = -1 \): 5 values
- For \( a = -2 \): 3 values

Thus, for the second inequality, we have \( 5 + 5 + 3 + 5 + 3 = 21 \) pairs.

### Step 3: Analyze \( a^2 + b^2 < 8b \)

Rearranging gives \( a^2 + b^2 - 8b < 0 \) or \( a^2 + (b - 4)^2 < 16 \). This describes a circle centered at (0, 4) with radius 4.

1. **For \( b = 0 \)**:
- \( a^2 + 16 < 16 \) → No valid values for \( a \).

2. **For \( b = 1 \)**:
- \( a^2 + 9 < 16 \) → \( a \) can be -3, -2, -1, 0, 1, 2, 3 (7 values).

3. **For \( b = 2 \)**:
- \( a^2 + 4 < 16 \) → \( a \) can be -3, -2, -1, 0, 1, 2, 3 (7 values).

4. **For \( b = 3 \)**:
- \( a^2 + 1 < 16 \) → \( a \) can be -3, -2, -1, 0, 1, 2, 3 (7 values).

5. **For \( b = 4 \)**:
- \( a^2 < 0 \) → No valid values for \( a \).

Calculating the total:
- For \( b = 1 \): 7 values
- For \( b = 2 \): 7 values
- For \( b = 3 \): 7 values

Thus, for the third inequality, we have \( 7 + 7 + 7 = 21 \) pairs.

### Step 4: Combine the results

The ordered pairs must satisfy all three inequalities. The most restrictive condition is \( a^2 + b^2 < 8 \), which gives us 21 pairs.

### Final Answer

Thus, the number of ordered pairs of integers \( (a, b) \) that satisfy all the given inequalities is **21**.