The number of integral solutions of `y^4=2x^4+1402.`

Options

A. 4
B. 2
C. 1
D. none

Answer

To find the number of integral solutions for the equation \( y^4 = 2x^4 + 1402 \), we will follow these steps:

### Step 1: Rearranging the Equation
We start by rearranging the equation:
\[ y^4 - 2x^4 = 1402 \]

### Step 2: Factoring the Left Side
We can factor the left-hand side using the difference of squares:
\[ (y^2 + \sqrt{2}x^2)(y^2 - \sqrt{2}x^2) = 1402 \]

### Step 3: Finding Factor Pairs of 1402
Next, we need to find the factor pairs of 1402. The prime factorization of 1402 is:
\[ 1402 = 2 \times 701 \]
Since 701 is a prime number, the factor pairs of 1402 are:
1. \( (1, 1402) \)
2. \( (2, 701) \)

### Step 4: Setting Up Equations from Factor Pairs
For each factor pair \( (a, b) \), we set up the equations:
1. \( y^2 + \sqrt{2}x^2 = a \)
2. \( y^2 - \sqrt{2}x^2 = b \)

### Step 5: Solving for \( y^2 \) and \( x^2 \)
From the two equations, we can add and subtract to find \( y^2 \) and \( \sqrt{2}x^2 \):
\[ y^2 = \frac{a + b}{2} \]
\[ \sqrt{2}x^2 = \frac{a - b}{2} \]

### Step 6: Checking for Integral Solutions
Now we will check each factor pair to see if \( y^2 \) and \( x^2 \) yield integral values.

1. For the pair \( (1, 1402) \):
\[ y^2 = \frac{1 + 1402}{2} = \frac{1403}{2} \quad \text{(not an integer)} \]
\[ \sqrt{2}x^2 = \frac{1 - 1402}{2} = \frac{-1401}{2} \quad \text{(not an integer)} \]

2. For the pair \( (2, 701) \):
\[ y^2 = \frac{2 + 701}{2} = \frac{703}{2} \quad \text{(not an integer)} \]
\[ \sqrt{2}x^2 = \frac{2 - 701}{2} = \frac{-699}{2} \quad \text{(not an integer)} \]

### Conclusion
Since neither factor pair yields integral values for \( y^2 \) and \( x^2 \), we conclude that there are no integral solutions for the equation \( y^4 = 2x^4 + 1402 \).

**Final Answer: There are no integral solutions.**

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