For any Fermat number $F_n=2^{2^n}+1$ with $n>0$, establish that $F_n\equiv5\,\text{or}\,8\pmod9$ For any Fermat number $F_n=2^{2^n}+1$ with $n>0$, establish that $F_n\equiv5\,\text{or}\,8\pmod9$ according as $n$ is odd or even.
Since for any $n\ge3$,$2^{2^n}\equiv2^{2^{n-2}}\pmod9$, if $n=2k+3$
then
$$2^{2^n}=2^{2^{2k+3}}\equiv2^{2^{2k+1}}=(2^{2^{2k}})^2\pmod9?$$
if $n=2k+4$
then
$$2^{2^n}=2^{2^{2k+4}}\equiv2^{2^{2k+2}}=(2^{2^{2k}})^4\pmod9?$$
but I don't know how continue?

Note that $2^6 \equiv 1 \pmod 9$. 
Also $2^n \equiv 2,4 \pmod 6$ depending on the parity of $n$. If $n$ is odd then $2^n \equiv 2 \pmod 6$, otherwise $2^n  \equiv 4 \pmod 6$. This means that $2^n = 6k + 2$ or $2^n = 6k + 4$. Now just substitute and use the first conclusion.
