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.
PARITY et all
