December 21st, 2013, 6:06 am
how about proving it for p=1729 (the "dull" Hardy-Ramanujan #)? and how about p=286 or 24046?----- ----- ----- ----- -----spoiler/comments:(1) simply expand the powers using binomial theorem, one gets 3^(p-1)=1 (mod p). besides all primes (as indicated above by Fermat's little theorem), there are composite numbers that satisfy this relation: Fermat pseudo-primes to base 3;(2) one might consider the general form (p+m)^(p+n)-(p+n)^(p+m)=0 (mod p). if sufficiently large prime p satisfies the equation, seems one needs either m or n=-1 or m=-n