1
$\begingroup$

How do I achieve this result via Pepin's test by using Euler's Theorem and such to simplify $3^{2^{31}}$ and get the desired congruence of 10,324,303?

$3^{(F_5-1)/2} = 3^{2^{31}} = 3^{2,146,483,648} \equiv 10,324,303 ≢ -1 (mod 4,294,967,297)$

Where $F_5$ is Fermat's 6th number $2^{32}+1$

Using powermod functions gives the desired result for $a^{b} mod(m)$ but I'd like to do this one by hand to better understand how powermod simplifies.

Thanks!

  • 0
    Do you really want to calculate it '**by hand**'?? What is the *real* question you want to ask?2012-09-19
  • 0
    If you want to use Euler's Theorem, you will need the factorization of $F_5$ in order to find $\varphi(F_5)$. The factorization is known, but rather goes against the spirit of Pepin's Test.2012-09-19

1 Answers 1