Why $2^{3^n}=-1 \mod 3^{n+1}?$ Why $2^{3^n}=-1 \mod 3^{n+1}?$
By using the Euler totient function $\varphi$   I have got that
$$
2^{\varphi(3^{n+1})}=2^{3^{n+1}-3^n}=2^{2 \cdot  3^n}=1 \mod 3^{n+1}.
$$
How  to prove now  that the square  root is equal  to $-1?$

We have $$(2^{3^n}+1)(2^{3^n}-1) = 0 \mod{3^{n+1}}$$ 
then remark that $2^{3^n }- 1 = (3-1)^{3^n} - 1 = (-1)^{3^n} - 1 = -2 = 1 \mod{3}$
so $$2^{3^n}+1 = 0 \mod{3^{n+1}}$$ 
