Understanding the Prime Number Problem

The question asks us to find the number of possible combinations for \((x, y, x+y, x - y)\) given specific conditions. Let's break down these conditions:

• \(x\) and \(y\) must be natural numbers. Natural numbers are positive whole numbers: \(\{1, 2, 3, ...\}\).
• Both \(x\) and \(y\) must be less than 20. So, \(1 \le x \le 19\) and \(1 \le y \le 19\).
• The numbers \(x\), \(y\), \(x+y\), and \(x-y\) must all be prime numbers.

First, let's list the prime numbers less than 20:

Primes < 20 = \(\{2, 3, 5, 7, 11, 13, 17, 19\}\)

Since \(x-y\) must be a prime number, it must be positive. This means \(x\) must be greater than \(y\) (\(x > y\)).

Analyzing the Conditions for x and y

We need \(x\) and \(y\) to be prime numbers themselves, and \(x > y\). The possible pairs of \((x, y)\) where both are primes less than 20 and \(x > y\) are:

• \(y=2\), \(x \in \{3, 5, 7, 11, 13, 17, 19\}\)
• \(y=3\), \(x \in \{5, 7, 11, 13, 17, 19\}\)
• \(y=5\), \(x \in \{7, 11, 13, 17, 19\}\)
• \(y=7\), \(x \in \{11, 13, 17, 19\}\)
• \(y=11\), \(x \in \{13, 17, 19\}\)
• \(y=13\), \(x \in \{17, 19\}\)
• \(y=17\), \(x \in \{19\}\)

Now let's consider the parity (even or odd) of \(x\) and \(y\). Remember, the only even prime number is 2.

• Case 1: \(y\) is an odd prime.
  If \(y\) is an odd prime (like 3, 5, 7, etc.), then \(x\) must also be an odd prime (since \(x > y\) and \(x\) must be prime).
  • If both \(x\) and \(y\) are odd primes, then \(x-y\) (odd - odd) would be an even number. The only even prime is 2. So, \(x-y = 2\).
  • Similarly, \(x+y\) (odd + odd) would be an even number. For \(x+y\) to be prime, it must be 2. However, since \(x > y \ge 3\), \(x+y\) must be greater than \(3+3=6\). Thus, \(x+y\) cannot be 2.
  • Therefore, it's impossible for both \(x\) and \(y\) to be odd primes.
• Case 2: \(y\) is the even prime.
  The only possibility is \(y=2\). Since \(x > y\), \(x\) must be an odd prime.
  • We need \(x\) (odd prime < 20), \(y=2\) (prime), \(x+y\) (prime), and \(x-y\) (prime).
  • Let's check the possible values for \(x\) from the list of odd primes greater than 2 and less than 20: \(x \in \{3, 5, 7, 11, 13, 17, 19\}\).

Testing Potential Combinations

We will test each possible value of \(x\) (where \(y=2\)):

• If \(x=3, y=2\):
  • \(x=3\) (prime), \(y=2\) (prime)
  • \(x+y = 3+2 = 5\) (prime)
  • \(x-y = 3-2 = 1\) (not prime)
  This combination fails because \(x-y\) is not prime.
• If \(x=5, y=2\):
  • \(x=5\) (prime), \(y=2\) (prime)
  • \(x+y = 5+2 = 7\) (prime)
  • \(x-y = 5-2 = 3\) (prime)
  This combination works! \(x=5\), \(y=2\), \(x+y=7\), \(x-y=3\). All are primes, and \(x, y < 20\). The combination is \((5, 2, 7, 3)\).
• If \(x=7, y=2\):
  • \(x=7\) (prime), \(y=2\) (prime)
  • \(x+y = 7+2 = 9\) (not prime)
  • \(x-y = 7-2 = 5\) (prime)
  This combination fails because \(x+y\) is not prime.
• If \(x=11, y=2\):
  • \(x=11\) (prime), \(y=2\) (prime)
  • \(x+y = 11+2 = 13\) (prime)
  • \(x-y = 11-2 = 9\) (not prime)
  This combination fails because \(x-y\) is not prime.
• If \(x=13, y=2\):
  • \(x=13\) (prime), \(y=2\) (prime)
  • \(x+y = 13+2 = 15\) (not prime)
  • \(x-y = 13-2 = 11\) (prime)
  This combination fails because \(x+y\) is not prime.
• If \(x=17, y=2\):
  • \(x=17\) (prime), \(y=2\) (prime)
  • \(x+y = 17+2 = 19\) (prime)
  • \(x-y = 17-2 = 15\) (not prime)
  This combination fails because \(x-y\) is not prime.
• If \(x=19, y=2\):
  • \(x=19\) (prime), \(y=2\) (prime)
  • \(x+y = 19+2 = 21\) (not prime)
  • \(x-y = 19-2 = 17\) (prime)
  This combination fails because \(x+y\) is not prime.

Conclusion on Combinations

By systematically checking all possibilities based on the properties of prime numbers and the given conditions, we found only one pair \((x, y)\) that satisfies all requirements:

• \((x, y) = (5, 2)\)

This leads to the combination \((x, y, x+y, x-y) = (5, 2, 7, 3)\). All numbers in this tuple are prime, and \(x=5\) and \(y=2\) are natural numbers less than 20.

Therefore, there is only one such combination possible.