Show that 41 is a Germain prime.
Explain why $(2 × 41) + 1 = 83$ is not a factor of $2^{41}− 1$.
41 is a Germain prime if 41 is prime, and $2(41)+1$ is prime.
41 is prime. $2(41)+1=83$ is also prime.
And so 41 is a Germain prime.
Corollary (4.21) states that if Germain prime $q \equiv 1 \pmod 4$ then $2q+1 \mid 2^q+1$.
In our case, $41 \equiv 1 \pmod 4$, and so $83 \mid 2^{41}+1$.
This means, for some integer $k$
$$ \begin{align} 83k & = 2^{41}+1 \\ \\ 83k-2 & = 2^{41}-1 \\ \\ 83(k-\frac{2}{83}) & = 2^{41}-1 \end{align} $$
This shows that 83 is not a factor of $2^{41}-1$, because there is no integer $k$ for which $(k-\frac{2}{83})$ is an integer.