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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.

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    $\begingroup$ Well, what do you get when you raise both sides to that power? Try writing it out. $\endgroup$ – Bob Jones Mar 8 '17 at 3:45
  • $\begingroup$ @BobJones would I get $2^{2^{m^{2^{2^m-m}}}}\equiv-1^{2^{2^m-m}}($mod $F_m)$? The only sense that I can make out of that is that $-1^2=1$, and that $2^{2^2}=16$. $\endgroup$ – Ephraim Mar 8 '17 at 3:52
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    $\begingroup$ You do not take the first $m$ to the power. You know the rule $(a^b)^c=a^{b\cdot c}$, right? Let $a=2, b=2^m, c=2^{2^m-m}$. $\endgroup$ – Bob Jones Mar 8 '17 at 3:55
  • $\begingroup$ @BobJones Thank you, that was helpful. I believe that leaves us with $2^{2^{2^m}}=16^m\equiv1^{2^m-m}($mod $F_m$). $\endgroup$ – Ephraim Mar 8 '17 at 4:07
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$F_m=2^{2^m}+1$ 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!

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    $\begingroup$ The left side is unfortunately not $2^{F_m}$ as you need. Instead of adding $1$, you need to multiply by $2$. But it works out the same. $\endgroup$ – Bob Jones Mar 8 '17 at 4:46

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