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.
 A: Hint: 
$$2^{3x+1} = 8^x\times 2$$
$$8 \equiv 1\pmod{7} \implies 8^x \equiv 1 \pmod{7}\implies 8^x \times 2\equiv 2\pmod{7}.$$
A: For natural $x$ we have
$$4^{3x+1}+2^{3x+1}=4\cdot64^x+2\cdot8^x+1=4(64^x-1)+2(8^x-1)+7\equiv0\pmod{7}.$$
A: $X:=2^{3x+1}=2\cdot (7+1)^x \equiv 2\pmod{7}$. Hence
$$
X^2+X+1 \equiv 2^2+2+1 \equiv 0\pmod{7}.
$$
A: Attempt at induction.
$\begin{array}\\
4^{3(x+1)+1} + 2^{3(x+1)+1}
&=4^{3x+1+3} + 2^{3x+1+3}\\
&=4^34^{3x+1} + 2^32^{3x+1}\\
&=64\cdot 4^{3x+1} + 8\cdot 2^{3x+1}\\
&=(63+1)\cdot 4^{3x+1} + (7+1)\cdot 2^{3x+1}\\
&=63\cdot 4^{3x+1} + 7\cdot 2^{3x+1}+4^{3x+1}+2^{3x+1}\\
&=7(9\cdot 4^{3x+1} + 2^{3x+1})+4^{3x+1}+2^{3x+1}\\
&\equiv 4^{3x+1}+2^{3x+1}\pmod{7}\\
\end{array}
$
Therefore, if
$4^{3x+1} + 2^{3x+1} + 1
\equiv 0 \pmod{7}
$
then
$4^{3(x+1)+1} + 2^{3(x+1)+1} + 1
\equiv 0 \pmod{7}
$.
Since
$4^{3x+1} + 2^{3x+1} + 1
\equiv 0 \pmod{7}
$
for $x = 0$,
it is true for all $x$.
A: 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 $3\nmid m$
A: $$4^3\equiv 2^3\equiv 1\pmod{7} \tag{A}$$
$$ 4^{3x}\equiv 2^{3x}\equiv 1\pmod{7}\tag{B}$$
$$ 4^{3x+1}+2^{3x+1}+1\equiv 4+2+1 \equiv 0\pmod{7}.\tag{C}$$
