<s>[INST] Show that $4^{3x+1} + 2^{3x+1}+ 1$ is divisible by 7 I want to show that $4^{3x+1} + 2^{3x+1} + 1$ is divisible by 7, I am trying to show this with modular arithmetic. 
If I break up each part of the equation, I can see that 
$4^{3x+1} = 4$ x $2^{6x}$ 
which implies that $4^{3x+1}mod(7) = 4$
I can't quite find a nice factorization of $2^{3x+1}$
Any help specifically on how to treat $2^{3x+1}$ would be appreciated.
 [/INST] For natural $x$ we have
$$4^{3x+1}+2^{3x+1}=4\cdot64^x+2\cdot8^x+1=4(64^x-1)+2(8^x-1)+7\equiv0\pmod{7}.$$
</s>