<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] $$4^3\equiv 2^3\equiv 1\pmod{7} \tag{A}$$
$$ 4^{3x}\equiv 2^{3x}\equiv 1\pmod{7}\tag{B}$$
$$ 4^{3x+1}+2^{3x+1}+1\equiv 4+2+1 \equiv 0\pmod{7}.\tag{C}$$
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