Prove that there exists exactly one prime of the form n^3 −1.

Answer 1

let n= 1, 1^3 -1 = 0 is not prime

let n=2 2^3 -1 = 7 is prime

let n=3 3^3 -1 = 26 is not prime

let n=4, 4^3 - 1 = 63 is divisible by 9 and is not prime

let n=6, 6^3 -1 = 215, divisible by 5 and not prime

let n= 8 8^3 -1 = 511 is divisible by 7 and not prime

let n= 10, 10^3 -1 = 999, divisible by 9, not prime

let n=12, 12^3 -1 = 1727, divisible by 11, not prime

let n= 14, 14^3 - 1 = 2743, divisible by 13, not prime

let n= 16 16^3 -1 = 4095, divisible by 3,5, and 15, not prime

let n= 18, 18^3-1 = 5831, divisible by 17

let n= 20, 20^3 - 1 = 7999 divisible by 19

22, divisible by 21

24, divisible by 23

26, divisible by 25

28, divisible by 27

30, divisible by 29

n, divisible by n-1

any odd n will have n^3 -1 = an even number which is divisible by 2 and not prime

any even n will have n^3 -1 = (n-1)(n^2+n+1) and be divisible by n-1, but if n=2, being divisible by 1 does not make it non-prime. if n=2, then n^3-1 = 7 which is the only prime