<s>[INST] Prove that $2^{3n+1} + 5$ is a multiple of 7 for all n ≥ 0. As the title states I need to prove that $(2^{3n+1}+5)$ is a multiple of 7 for all $n \geq 0$.  
I can do this using induction but I also want to prove it using modular arithmetic. So here's what I've got so far. 
Start with Fermat's little theorem: 
\begin{align}2^6 &\equiv 1 \ \ \pmod{7} \\ 2^3 \cdot 2^3 &\equiv 1 \ \ \pmod{7} \\8 \cdot 2^3 &\equiv 1 \ \ \pmod{7}  \\ 2^3 &\equiv 1 \ \ \pmod{7} \\ 2^{3n} &\equiv 1^n \pmod{7} \end{align}
I feel up to this point everything is correct. 
How would I add the $1$ to the exponent? 
\begin{align} 2^{3n+1} &\equiv 1^n (5)  \pmod{7} \\ 2^{3n+1} &\equiv 5 \ \ \ \ \ \ \ \pmod{7}
\end{align}
Now adding the $5$, I am confused as to how to do that as well. I would just subtract the $5$ remainder correct? Such that: 
$2^{3n+1}  -5 \equiv 0 \pmod{7}$   but this is not what I intend to do. 
I need,  $2^{3n+1} +5 \equiv 0 \pmod{7}$
 [/INST] When you add $1$ to the exponent on $2$ you multiply by $2$, not by $5$ or $-2$.  So $2^{3n+1}\equiv 1×2\bmod 7$ and the rest is easy.
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