<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] 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 $x$.
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