If $p\mid (3^n+1)$, then $p\equiv 1\pmod{3}$ 
Show that if $p> 2$ is a prime, $n > 1$ is odd and $p\mid (3^n+1)$,
  then $p\equiv 1\pmod{3}$.

Since $n$ is odd, we have $3^{n+1} \equiv -3 \pmod{p}$ is a quadratic residue. Then I thought about using Quadratic Reciprocity but didn't see how to apply it. 
We have $x^2 \equiv -3 \pmod{p}$ for some integer $x$. By Fermat's Little Theorem we have $$x^{p-1} \equiv x^2 \cdot x^{p-3} \equiv -3 \cdot x^{p-3} \equiv 1 \pmod{p}.$$ Thus, $x^{p-3} \equiv -3^{-1} \pmod{p}$. 
I didn't see how to get a contradiction if $p \equiv 2 \pmod{3}$.

This may qualify as cheating, but I will use a few known facts from Wikipedia. First of all, indeed $3^{n+1} \equiv -3 \pmod{p}$, where $n+1$ is even, thus 
$$\left(\frac{-3}{p}\right)=1 \Rightarrow \left(\frac{-1}{p}\right)\left(\frac{3}{p}\right)=1 \tag{1}$$
But
$$\left(\frac{-1}{p}\right)=\left\{\begin{matrix}
1, & p \equiv 1 \pmod{4} \\ 
-1,& p \equiv 3 \pmod{4}  
\end{matrix}\right. \color{red}{\text{ and }} \left(\frac{3}{p}\right)=\left\{\begin{matrix}
1, & p \equiv 1 \text{ or } 11 \pmod{12} \\ 
-1, & p \equiv 5 \text{ or } 7 \pmod{12} 
\end{matrix}\right.$$
Due to $(1)$, we have 
$$\left\{\begin{matrix}
\left(\frac{-1}{p}\right)=1\\ 
\left(\frac{3}{p}\right)=1 
\end{matrix}\right. \color{red}{\text{ or }} \left\{\begin{matrix}
\left(\frac{-1}{p}\right)=-1\\ 
\left(\frac{3}{p}\right)=-1 
\end{matrix}\right. $$
Or, for the first case
$$\left\{\begin{matrix}
\left(\frac{-1}{p}\right)=1\\ 
\left(\frac{3}{p}\right)=1 
\end{matrix}\right. \Leftrightarrow \left\{\begin{matrix}
p \equiv 1 \pmod{4}\\ 
p \equiv 1 \text{ or } 11 \pmod{12} 
\end{matrix}\right. \Rightarrow p \equiv 1 \pmod{12}$$
and for the latter 
$$\left\{\begin{matrix}
\left(\frac{-1}{p}\right)=-1\\ 
\left(\frac{3}{p}\right)=-1 
\end{matrix}\right. \Leftrightarrow \left\{\begin{matrix}
p \equiv 3 \pmod{4}\\ 
p \equiv 5 \text{ or } 7 \pmod{12} 
\end{matrix}\right. \Rightarrow p \equiv 7 \pmod{12}$$
As a result 
$$p \equiv 1 \pmod{12} \color{red}{\text{ or }} p \equiv 7 \pmod{12} \Rightarrow \\ p \equiv 1 \pmod{3} \color{red}{\text{ or }} p \equiv 7 \pmod{3} \Rightarrow \\ p \equiv 1 \pmod{3}$$
