Fermat's little theorem states that if $p$ is prime, then $\forall a\in\mathbb N \quad a^p\equiv a \quad \text{mod} \quad p$. Is the converse true ?
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4https://en.wikipedia.org/wiki/Carmichael_number – Angina Seng Dec 20 '19 at 20:03
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If memory serves, you might want to look up Carmichael numbers. They act like primes, but are not. – Dec 20 '19 at 20:06
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1See also Fermat pseudoprimes for the converse notion for fixed $a$. – Arthur Dec 20 '19 at 20:10
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No, carmichael numbers are counterexamples, e.g. see here and here. But there are some valid comverses e.g that by Lucas – Bill Dubuque Dec 20 '19 at 21:09