<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] $X:=2^{3x+1}=2\cdot (7+1)^x \equiv 2\pmod{7}$. Hence
$$
X^2+X+1 \equiv 2^2+2+1 \equiv 0\pmod{7}.
$$
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