19 is the only prime which is equal to the difference of two cubes of primes.
Prove the above statement.
Let p and q, with p>q, be two prime numbers, such that:
p^3-q^3 = 19
=> (p-q)(p^2+pq+q^2) = 19
Since each of p and a is a prime number, each of them must be > 1, so that:
p^2+pq+q^2 is also >1
In this situation, if p-q>1, then p^3-q^3 is a composite number. This is a contradiction.
Therefore, we must have: p-q=1
Hence, one of p and q must be even and the other must be odd.
Since each of p and q is a prime number, this is possible only when p=3 and q=2.
Substituting these values, we observe that: 3^3-2^3=19
Consequently, 19 is the only prime which is equal to the difference of two cubes of primes.
Puzzle Solution
