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Find all pairs of prime numbers, the difference of the squares of which is a prime number.

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Let $p$ and $q$ be prime numbers and $p^{2}-q^{2}=(p-q)(p+q)$ be a prime number. Then $p-q=1$. Therefore, one of our numbers is even, that is, $q=2, p=3$.

## Answer

$(3,2)$.