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I know that Fermat Pseudoprime formula is $a^{p-1} \equiv 1 \mod{p}$ c but in this case $p$ is not prime.

How do I prove that that $91$ is/is not a pseudoprime to base $2$.

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$91$ is not prime , but $91=7*13$. Moreover, $$ 2^{90}= 2^{6*15}=64^{15} \equiv (-1)^{15} \equiv -1 \mod{13} $$ since $13 *5 =65$. Hence $13 \nmid 2^{91-1 }-1$ as $ 13 \nmid 2^{91-1 }+1 $ and so $91 \nmid 2^{91-1 } $. Therefore $91$ is not Fermat Pseudoprime of base 2.

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