32.1. Find all such prime numbers that can be represented both as the sum of two prime numbers and as their difference. $(7-10$ grades $)$
32.1. It is obvious that neither of the prime numbers 2 and 3 can be represented as the sum of two prime numbers. Now, let's take a prime number \( p > 3 \). It can be represented as the sum of two prime numbers only if \( p-2 \) is a prime number, since in the sum of two prime numbers, one of the addends must be equal to 2. Similarly, if a prime number \( p > 3 \) can be represented as the difference of two prime numbers, then the number \( p+2 \) must be prime, since in the difference of two prime numbers, the subtrahend must be equal to 2. Therefore, if a prime number \( p \) can be represented both as the sum and as the difference of two prime numbers, then the numbers \( p-2 \) and \( p+2 \) must be prime simultaneously. It is clear that the number 5 satisfies the condition of the problem, since \( 5 = 3 + 2 = 7 - 2 \), where all four numbers \( 2, 3, 5, 7 \) are prime. We will prove that the problem has no other solutions. For this, it is sufficient to prove that for a prime \( p > 5 \), the numbers \( p-2 \) and \( p+2 \) cannot be prime simultaneously. Indeed, a prime number \( p > 5 \) when divided by 3 gives a remainder of 1 or 2, i.e., it has one of two forms: \( p = 3k + 1 \) or \( p = 3k + 2 \), where \( k > 1 \). In the first case, the number \( p + 2 = 3k + 1 + 2 = 3(k + 1) \) will be composite, and in the second case, the number \( p - 2 = 3k + 2 - 2 = 3k \) will be composite.