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Let $n$ be a Fermat-pseudoprime with respect to 2. Show that $2^n-1$ is a strong pseudoprime with respect to 2.

I tried solving the problem as follows. Since $n$ is a Fermat-pseudoprime with respect to 2, it holds that $2^{n-1}\equiv 1\mod n$, so $n$ divides $2^{n-1}-1$. Therefore, since $2^n\equiv1\mod 2^n-1$, we have, by raising both sides of the equivalence to a suitable power, that $2^{2^{n-1}-1}\equiv 1\mod2^n-1$. This implies exactly that $2^n-1$ is a strong pseudoprime with respect to 2, since $(2^n-1)-1=2(2^{n-1}-1)$.

Is this attempt correct? Also, how do I prove that $2^n-1$ is not a prime?

bitsbit
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1 Answers1

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It is correct. Suppose $d$ divides $n$, since $n$ is assumed to be nonprime. Then $2^d-1$ divides $2^n-1$, so $2^n-1$ is not prime.

littlecat
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