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

Options

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

Answer

To find the number of integer solutions \((a, b)\) for the equation \(2a^2 + 3b^2 = 35\), we will follow these steps:

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

### Step 2: Finding Integer Values for \(a\)
Since \(b^2\) must be a non-negative integer, \(35 - 2a^2\) must also be non-negative. Therefore, we need:
\[ 35 - 2a^2 \geq 0 \implies 2a^2 \leq 35 \implies a^2 \leq 17.5 \]
This means \(a^2\) can take values from \(0\) to \(17\). The possible integer values for \(a\) are:
\[ a = -4, -3, -2, -1, 0, 1, 2, 3, 4 \]

### Step 3: Calculating Corresponding Values of \(b\)
Now we will substitute each integer value of \(a\) back into the equation to find corresponding \(b\) values.

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

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

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

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

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

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

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

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

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

### Step 4: Collecting All Solutions
From the calculations, we have the following integer solutions:
- \((2, 3)\)
- \((2, -3)\)
- \((4, 1)\)
- \((4, -1)\)
- \((-2, 3)\)
- \((-2, -3)\)
- \((-4, 1)\)
- \((-4, -1)\)

### Conclusion
Thus, the total number of integer solutions \((a, b)\) is \(8\).

### Final Answer
The number of elements in the set \(\{(a,b): 2a^2 + 3b^2 = 35, a,b \in \mathbb{Z}\}\) is \(8\).

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