The number 3n+4n is odd and not divisible by 3, therefore n is odd as well and not divisible by 3.
Write n=2k+1. Then
3n+4n=3.9k+42k+13.(-1)k-1 (mod 5).
This number is not divisible by 5 for any value of n, so neither 5 is a divisor of n.
Let p be the smallest prime divisor of n and c is that residue class for which 4c3 (mod p). Then
4n(c2n-1)=(cn-1)((4c)n+4n)(cn-1)(3n+4n)0 (mod p)