2
$\begingroup$

I have to prove that $n = \frac{4^p - 1}{3}$ is a Fermat pseudoprime with respect to $2$ when $p \geq 5$ is a prime number. I have proved that $n$ is not prime because $4^p - 1 = (2^p-1)(2^p+1)$ and $(2^p + 1)$ is divisible by $ 3$. But now I can't show that $2^{n-1} \equiv 1\bmod n$.

I calculated that $2^{n-1} = 2^{(2^p + 2)(2^p-2)/3}$ but I don't know if I can deduce anything from this.

$\endgroup$
4
$\begingroup$

$n=\dfrac{4^p-1}3=\dfrac{2^{2p}-1}3,$ so $n$ divides $2^{2p}-1$.

Furthermore, $2p$ divides $2\times\dfrac{(2^{p-1}-1)}3\times{(2^{p}+2)}=\dfrac{2^{2p}-4}3=n-1.$

Therefore, $n$ divides $2^{n-1}-1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.