Find all triples (x, y, z) of positive integers such that
x³ + y³ + z³ - 3xyz = p

where p is a prime greater than 3

Upon parity considerations, it was clear to me that the triplet must contain two even and one odd integer, otherwise p=even.

Although I have found few sample triplets, I did not perceive

that there are two equal integers and one odd neighbor.

After seeing Ch's output I realized that the sums cover all prime numbers.

Enter SH, - Saying - "give me a prime (over 3) and I will give you a triplet" 17=>6,6,5, 19=>6,6,7, 101=>34,34,33 etc

**Nice puzzle - I have rated it 4.**