Question: Show that every composite Fermat number $F_m=2^{2^m}+1$ is a pseudoprime base 2.
Hint: Raise the congruence $2^{2^m}\equiv-1($mod $F_m)$ to the $2^{2^m-m}$th power.
Even with the hint, I'm fairly lost. I understand that $2^{2^m}+1\equiv0($mod $ F_m)$, so clearly it follows that $2^{2^m}\equiv-1($mod $F_m)$. I also know that for a positive integer $b$, if $n$ is a composite positive integer and $b^n\equiv b($mod $n)$ then $n$ is called a pseudoprime to the base $b$.
So, we want to show that $2^{F_m}\equiv2($mod $F_m)$ I don't understand how the hint helps us do this.
Thank you for your help, I'm happy to answer any questions that come up.
$F_m=2^{2^m}+1$ implies $2^{2^m}+1\equiv0($mod $ F_m)$, and it follows that $2^{2^m}\equiv-1($mod $F_m)$.
Raising both sides by $2^{2^m-m}$ gives us
$2^{2^m*2^{2^m-m}}=2^{2^{m+{2^m-m}}}=2^{2^{2^m}}\equiv-1^{2^{2^m-m}}=1^{2^m-m}=1^{2^m}/1^m=1^m/1^m=1($mod $F_m)$
or
$2^{2^{2^m}}\equiv1($ mod $F_m)$
multiplying both sides by $2$ leaves us with:
$2^{2^{2^m}+1}\equiv2($ mod $F_m)$ as was to be shown.
Thanks to Bob Jones for his help!