OFFSET
1,1
COMMENTS
n^5+1 can only be prime when n = 1, n^5+1 = 2. This is because of the polynomial factorization n^5+1 = (n+1) * (n^4 - n^3 + n^2 - n + 1) = (n+1)*A060884(n). Hence after the initial n=1 prime, the binomial can at best be semiprime and that only when both (n+1) and (n^4 - n^3 + n^2 - n + 1) are primes.
LINKS
Robert Price, Table of n, a(n) for n = 1..1414
FORMULA
a(n)^5 + 1 is semiprime. a(n)+1 is prime and a(n)^4 - a(n)^3 + a(n)^2 - a(n) + 1 is prime.
EXAMPLE
n n^5+1 = (n+1) * (n^4 - n^3 + n^2 - n + 1)
2 33 = 3 * 11
10 100001 = 11 * 9091
12 248833 = 13 * 19141
16 1048577 = 17 * 61681
MATHEMATICA
Select[Range[2600], PrimeOmega[#^5+1]==2&] (* Harvey P. Dale, May 20 2011 *)
Select[Range[200000], PrimeQ[# + 1] && PrimeQ[(#^5 + 1)/(# + 1)] &] (* Robert Price, Mar 09 2015 *)
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Jonathan Vos Post, Apr 02 2005
STATUS
approved