The number of pairs of positive integers (x,y) where x and y are prime numbers and `x^(2)-2y^(2)=1`, is

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the equation \( x^2 - 2y^2 = 1 \) for pairs of positive integers \( (x, y) \) where both \( x \) and \( y \) are prime numbers, we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ x^2 - 2y^2 = 1 \]
This can be rearranged to:
\[ x^2 = 2y^2 + 1 \]

### Step 2: Analyze the equation
Since \( x \) and \( y \) are both prime numbers, we need to find pairs of primes that satisfy this equation.

### Step 3: Check small prime numbers for \( y \)
Let's check small values of \( y \) (which are prime numbers) and see if we can find corresponding \( x \) values that are also prime.

- **Case 1:** Let \( y = 2 \):
\[ x^2 = 2(2^2) + 1 = 2 \cdot 4 + 1 = 8 + 1 = 9 \implies x = 3 \]
Here, \( (x, y) = (3, 2) \) is a valid pair.

- **Case 2:** Let \( y = 3 \):
\[ x^2 = 2(3^2) + 1 = 2 \cdot 9 + 1 = 18 + 1 = 19 \implies x = \sqrt{19} \]
\( \sqrt{19} \) is not an integer, so this pair does not work.

- **Case 3:** Let \( y = 5 \):
\[ x^2 = 2(5^2) + 1 = 2 \cdot 25 + 1 = 50 + 1 = 51 \implies x = \sqrt{51} \]
\( \sqrt{51} \) is not an integer, so this pair does not work.

- **Case 4:** Let \( y = 7 \):
\[ x^2 = 2(7^2) + 1 = 2 \cdot 49 + 1 = 98 + 1 = 99 \implies x = \sqrt{99} \]
\( \sqrt{99} \) is not an integer, so this pair does not work.

### Step 4: Continue checking larger primes
Continuing this process for larger prime numbers, we find that:

- For \( y = 11, 13, 17, \ldots \), the values of \( x^2 \) will yield non-square integers, and thus no valid prime \( x \) will be found.

### Conclusion
After checking all small prime numbers for \( y \) and finding corresponding \( x \) values, we conclude that the only valid pair of positive integers \( (x, y) \) where both are prime is \( (3, 2) \).

Thus, the number of pairs of positive integers \( (x, y) \) where both are prime numbers and satisfy the equation \( x^2 - 2y^2 = 1 \) is:
\[ \boxed{1} \]