For two values that differ by two, their values in the expression always have a common factor:
[(n+1)^44][(n1)^4+4]=8n^2+8n=8n(n^2+1)
This n^2+1 is a common factor:
13^4+4=5*29*197
15^4+4=197*259
14^2+1=197
So since a=2 and a=3 are not prime, not greater values of a can yield primes.

Posted by Jer
on 20160713 10:20:51 