If x and y are possible prime number and if `x^2 - 2y^2 = 1`, then the value of x + y is :

Options

A. 7
B. 24
C. 13
D. none of these

Answer

To solve the equation \( x^2 - 2y^2 = 1 \) where \( x \) and \( y \) are prime numbers, we will follow these steps:

### Step 1: Rearrange the equation
We start with the equation:
\[ x^2 - 2y^2 = 1 \]
Rearranging gives:
\[ x^2 = 2y^2 + 1 \]

### Step 2: Substitute possible prime values for \( y \)
Since \( y \) must be a prime number, we can try small prime numbers for \( y \). The smallest prime numbers are 2, 3, 5, etc.

### Step 3: Test \( y = 2 \)
Let's first try \( y = 2 \):
\[ x^2 = 2(2^2) + 1 = 2(4) + 1 = 8 + 1 = 9 \]
Taking the square root gives:
\[ x = 3 \]
Since 3 is also a prime number, this pair \( (x, y) = (3, 2) \) works.

### Step 4: Calculate \( x + y \)
Now we can calculate \( x + y \):
\[ x + y = 3 + 2 = 5 \]

### Step 5: Check for other possible values of \( y \)
Next, we can check for the next prime number, \( y = 3 \):
\[ x^2 = 2(3^2) + 1 = 2(9) + 1 = 18 + 1 = 19 \]
Taking the square root gives:
\[ x = \sqrt{19} \quad (\text{not an integer, so discard}) \]

Next, try \( y = 5 \):
\[ x^2 = 2(5^2) + 1 = 2(25) + 1 = 50 + 1 = 51 \]
Taking the square root gives:
\[ x = \sqrt{51} \quad (\text{not an integer, so discard}) \]

### Conclusion
The only pair of prime numbers that satisfies the equation \( x^2 - 2y^2 = 1 \) is \( (3, 2) \). Therefore, the value of \( x + y \) is:
\[ \boxed{5} \]