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.