Numbers of the form k^n*(k^n+1) + 1 with n > 0, k > 1 may be primes only if n has the form 3^j. When n is even k^(4*n) + k^(2*n) + 1 = (k^(2*n)+1)^2 - (k^n)^2 = (k^(2*n)+k^n+1)*(k^(2*n)-k^n+1) so composite. But why if n odd > 3 and not a power of 3 is k^n*(k^n+1) + 1 always composite?
Phi[3^(n+1),k] = k^(3^n)*(k^(3^n)+1)+1. When m <> 3^n in k^m*(k^m+1)+1, Phi[3m,k] < k^m*(k^m+1)+1 and is a divisor of it. -
Lei Zhou, Feb 09 2012
The prime number corresponding to the 10th term is a 587458-digit number. -
Lei Zhou, Jul 04 2014
x^(2*k) + x^k + 1 = (x^(3*k) - 1)/(x^k - 1) is the product over n dividing 3k but not dividing k of cyclotomic polynomials Phi(n). If k is a power of 3, n = 3k is the only such divisor and we have a single irreducible cyclotomic polynomial Phi(3k). Otherwise we have the product of more than one polynomial, with integer values > 1 for integer x > 1, and thus always composite numbers. -
Martin Becker, Jun 22 2021
