Find all prime numbers which can be presented as a sum of two primes and difference of two primes at the same time.
1. Let the prime be \( p \). Then we have \( p = a + b \) and \( p = c - d \), where \( a \), \( b \), \( c \), and \( d \) are also primes. Without loss of generality, assume \( a > b \).

2. First, consider the case \( p = 2 \). If \( p = 2 \), then \( 2 = a + b \) and \( 2 = c - d \). Since 2 is the only even prime, both \( a \) and \( b \) must be primes that sum to 2. The only such pair is \( a = 2 \) and \( b = 0 \), but 0 is not a prime number. Therefore, \( p \neq 2 \).

3. Since \( p \) is not 2, \( p \) must be an odd prime. For \( a + b \) to be odd, one of \( a \) or \( b \) must be even, and the other must be odd. The only even prime is 2, so without loss of generality, let \( b = 2 \). Thus, \( p = a + 2 \).

4. Similarly, for \( c - d \) to be odd, one of \( c \) or \( d \) must be even, and the other must be odd. The only even prime is 2, so without loss of generality, let \( c = 2 \). Thus, \( p = 2 - d \).

5. Now we have the equations:
   \[
   p = a + 2
   \]
   \[
   p = 2 - d
   \]

6. Equating the two expressions for \( p \), we get:
   \[
   a + 2 = 2 - d
   \]
   \[
   a + d = 0
   \]
   Since \( a \) and \( d \) are primes, the only solution to this equation is \( a = 0 \) and \( d = 0 \), but 0 is not a prime number. Therefore, we need to reconsider our approach.

7. Instead, consider the condition \( p = a + 2 \) and \( p = d - 2 \). This implies:
   \[
   a + 2 = d - 2
   \]
   \[
   d = a + 4
   \]

8. Since \( a \), \( p \), and \( d \) are all primes, and they form three consecutive odd numbers, one of them must be a multiple of 3 (since every third odd number is a multiple of 3).

9. The only set of three consecutive odd primes where one is a multiple of 3 is \( 3, 5, 7 \). Here, \( a = 3 \), \( p = 5 \), and \( d = 7 \).

10. Therefore, the only prime number that can be expressed as both the sum of two primes and the difference of two primes is \( \boxed{5} \).