Let $\ M\ $ be the set of the prime numbers $\ p\ $ such that $\ p\nmid 3^n+5^n+7^n\ $ for every positive integer $\ n\ $ , in short the set of the prime numbers that cannot divide $\ 3^n+5^n+7^n\ $.
Is $\ M\ $ an infinite set ?
For $\ p\ge 11\ $ , it is sufficient that $\ p\nmid 3^n+5^n+7^n\ $ for all $\ n\ $ with $\ 1\le n\le p-1\ $ because of Fermat's little theorem. Also, it is eay to see that $\ 2\ $ and $\ 7\ $ are in $\ M\ $ , but $\ 3\ $ and $\ 5\ $ are not. The first primes in $\ M\ $ are
gp > forprime(p=1,500,gef=0;for(k=1,p-1,if(Mod(3^k+5^k+7^k,p)==0,gef=1));if(gef==0,print1(p," ")))
2 7 17 23 29 37 41 43 67 73 79 89 101 107 109 127 131 139 151 173 197 199 211 227 251 257 263 277 293 313 347 349 353 367 379 421 439 443 449 479 499
gp >
Larger examples include $\ 10^4+7\ $ and $\ 10^5+3\ $. It would be nice if we could classify the primes in $\ M\ $ somehow, but I have no idea how this should be possible.
That infinite many primes are NOT in $\ M\ $ can be shown as follows :
For a given prime $\ q\ge 11\ $ , define $$N:=lcm(11-1,13-1,17-1,\cdots ,q-1)$$ Then for every prime $\ r\ $ with $\ 11\le r\le q\ $ we have $\ r-1\mid N\ $ and therefore $\ 3^N+5^N+7^N\equiv 3\ne 0\mod r\ $. Since $\ N\ $ is even , $\ 3^N+5^N+7^N\ $ cannot be divisible by $\ 3\ $ or $\ 5\ $. Hence, there is no prime factor smaller than or equal to $\ q\ $. This shows that the smallest prime factor can become arbitarily large.
