<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] If $x^2+x+1=0,x=w$ where $w$ is a complex cube root of unity.  
$$x^{2m}+x^m+1=w^{2m}+w^m+1= \begin{cases}(w^3)^{2n}+(w^3)^n+1=3 &\mbox{if }m=3n\\ 
w^{2(3n+1)}+w^{3n+1}+1=w^2+w+1=0 & \mbox{if }m=3n+1\\=\cdot=0 & \mbox{if }m=3n+2 \end{cases} $$
$\implies x^2+x+1$ will divide $x^{2m}+x^m+1$ if $3\nmid m$
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