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Find prime p, q, s.t. p^2 - 2q^2 = 1

Problem:
Find prime \(p, q\) which satisfies \(p^2 - 2q^2 = 1\).

Solution:
We see that \(p = 3, q = 2\) is a solution. We want to show that this is the only solution.

Suppose there exists a solution with \( q > 2 \). Rearranging the equation we have \( (p-1)(p+1) = 2q^2 \). Since RHS is divisible by 2, we know that \(p\) must be an odd prime. Hence, both \( p-1 \) and \( p+1 \) are divisible by 2. That means LHS is at least divisible by 4. Hence \( 2q^2 \) must be divisible by 4. This is only possible when \( 2 | q \). Contradiction.

QED

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