The numbers of integral solutions of the equations
`y^(2)(5x^(2)+1)=25(2x^(2)+13)" where "x, y inI`, is

Options

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

Answer

To solve the equation \( y^2(5x^2 + 1) = 25(2x^2 + 13) \) for integral solutions where \( x, y \in \mathbb{Z} \), we can follow these steps:

### Step 1: Rearranging the Equation
Start by rewriting the equation:
\[ y^2(5x^2 + 1) = 25(2x^2 + 13) \]
This can be rearranged as:
\[ y^2(5x^2 + 1) - 25(2x^2 + 13) = 0 \]

### Step 2: Isolating \( y^2 \)
We can express \( y^2 \) in terms of \( x \):
\[ y^2 = \frac{25(2x^2 + 13)}{5x^2 + 1} \]

### Step 3: Finding Conditions for \( y^2 \) to be an Integer
For \( y^2 \) to be an integer, the right-hand side must be an integer. Therefore, \( 5x^2 + 1 \) must divide \( 25(2x^2 + 13) \).

### Step 4: Simplifying the Division
We can simplify the expression:
\[ \frac{25(2x^2 + 13)}{5x^2 + 1} \]
Let’s perform polynomial long division or factorization to analyze the divisibility.

### Step 5: Finding Integer Solutions
We can check specific integer values for \( x \) to find corresponding \( y \).

1. **For \( x = 0 \)**:
\[ y^2 = \frac{25(2(0)^2 + 13)}{5(0)^2 + 1} = \frac{25 \cdot 13}{1} = 325 \quad \text{(not a perfect square)} \]

2. **For \( x = 1 \)**:
\[ y^2 = \frac{25(2(1)^2 + 13)}{5(1)^2 + 1} = \frac{25(2 + 13)}{5 + 1} = \frac{25 \cdot 15}{6} = 62.5 \quad \text{(not an integer)} \]

3. **For \( x = 2 \)**:
\[ y^2 = \frac{25(2(2)^2 + 13)}{5(2)^2 + 1} = \frac{25(8 + 13)}{20 + 1} = \frac{25 \cdot 21}{21} = 25 \quad \Rightarrow \quad y = \pm 5 \]

4. **For \( x = -2 \)** (since \( y^2 \) is symmetric):
\[ y^2 = \frac{25(2(-2)^2 + 13)}{5(-2)^2 + 1} = 25 \quad \Rightarrow \quad y = \pm 5 \]

5. **For \( x = 3 \)**:
\[ y^2 = \frac{25(2(3)^2 + 13)}{5(3)^2 + 1} = \frac{25(18 + 13)}{45 + 1} = \frac{25 \cdot 31}{46} \quad \text{(not an integer)} \]

6. **For \( x = -3 \)**:
\[ y^2 = \frac{25(2(-3)^2 + 13)}{5(-3)^2 + 1} = \frac{25(18 + 13)}{45 + 1} = \frac{25 \cdot 31}{46} \quad \text{(not an integer)} \]

### Step 6: Summary of Solutions
From the calculations, we find the valid pairs:
- \( (2, 5) \)
- \( (2, -5) \)
- \( (-2, 5) \)
- \( (-2, -5) \)

Thus, the total number of integral solutions is **4**.

### Final Answer
The number of integral solutions of the equation is **4**.