The number of elements in the set
`{(a, b): 2a^2 + 3b^2 = 35, a, b epsilon Z)`, where Z is the set of all integers, is

Options

A. 2
B. 4
C. 8
D. 12

Answer

To solve the problem of finding the number of elements in the set defined by the equation \(2a^2 + 3b^2 = 35\) where \(a\) and \(b\) are integers, we can follow these steps:

### Step 1: Rearranging the Equation
We start with the equation:
\[ 2a^2 + 3b^2 = 35 \]
We can rearrange this to isolate \(3b^2\):
\[ 3b^2 = 35 - 2a^2 \]

### Step 2: Finding the Range for \(b^2\)
Since \(b^2\) must be non-negative, we need \(35 - 2a^2 \geq 0\):
\[ 35 \geq 2a^2 \implies a^2 \leq \frac{35}{2} = 17.5 \]
Since \(a\) is an integer, the maximum integer value for \(a^2\) is 17, which means:
\[ a^2 \leq 17 \implies |a| \leq 4 \]
Thus, \(a\) can take the values \(-4, -3, -2, -1, 0, 1, 2, 3, 4\).

### Step 3: Finding Possible Values for \(b\)
Next, we will substitute possible integer values for \(a\) and find corresponding values for \(b\).

- **For \(a = 0\)**:
\[ 3b^2 = 35 \implies b^2 = \frac{35}{3} \text{ (not an integer)} \]

- **For \(a = \pm 1\)**:
\[ 2(1^2) + 3b^2 = 35 \implies 3b^2 = 33 \implies b^2 = 11 \implies b = \pm \sqrt{11} \text{ (not an integer)} \]

- **For \(a = \pm 2\)**:
\[ 2(2^2) + 3b^2 = 35 \implies 8 + 3b^2 = 35 \implies 3b^2 = 27 \implies b^2 = 9 \implies b = \pm 3 \]
Possible pairs: \((2, 3), (2, -3), (-2, 3), (-2, -3)\)

- **For \(a = \pm 3\)**:
\[ 2(3^2) + 3b^2 = 35 \implies 18 + 3b^2 = 35 \implies 3b^2 = 17 \implies b^2 = \frac{17}{3} \text{ (not an integer)} \]

- **For \(a = \pm 4\)**:
\[ 2(4^2) + 3b^2 = 35 \implies 32 + 3b^2 = 35 \implies 3b^2 = 3 \implies b^2 = 1 \implies b = \pm 1 \]
Possible pairs: \((4, 1), (4, -1), (-4, 1), (-4, -1)\)

### Step 4: Counting the Ordered Pairs
Now, we can summarize the valid pairs:
1. From \(a = 2\): \((2, 3), (2, -3), (-2, 3), (-2, -3)\) → 4 pairs
2. From \(a = 4\): \((4, 1), (4, -1), (-4, 1), (-4, -1)\) → 4 pairs

Total pairs = \(4 + 4 = 8\).

### Final Answer
The number of elements in the set is \(8\).

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