Find all triples (x, y, z) of positive integers such that
x³ + y³ + z³ - 3xyz = p
where p is a prime greater than 3
let f(x,y,z) =
x³ + y³ + z³ - 3xyz where x,y and z are positive integers
Then f(a,a,a+1) = 3a+1
f(a,a,a-1) = 3a-1
The minimum value of f is f(1,1,2) = 4
So, we can easily make the function equal any integer greater than 3 that is not divisible by 3. No primes greater than 3 are divisible by 3, so we can make the function equal any prime.
There are probably many other ways to calculate many primes, and probably even ways to make multiples of 3.
Edited on December 25, 2013, 2:27 pm
Charlie's proof
