If $a^n\equiv a \pmod n$ for all integers $a$, does this imply that $n$ is prime?
I believe this is the converse of Fermat's little theorem.
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4https://en.wikipedia.org/wiki/Carmichael_number – Will Jagy Feb 09 '16 at 03:37
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Try $n = 561$ or $n = 512461$. These are examples of Carmichael numbers (see the previous comment). The phenomenon is interesting enough to have attracted the attention of many well-known mathematicians. – Hans Engler Feb 09 '16 at 03:40
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1and see https://en.wikipedia.org/wiki/Fermat_primality_test#Flaw (but that test is still very useful, for example for RSA keys generation) – reuns Feb 09 '16 at 04:41
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1No, 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 21:13
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No, the converse of Fermat's Little Theorem is not true. For a particular example, $561 = 3 \cdot 11 \cdot 17$ is clearly composite, but $$ a^{561} \equiv a \pmod{561}$$ for all integers $a$. This is known as a Carmichael Number, and there are infinitely many Carmichael Numbers.
davidlowryduda
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Here is one more example:
We see that $341 = 11 \cdot31$ and $2^{340} = 1\mod341$
To show this we see that by routine calcultions the following relations hold $$2^{11} = 2\mod31 $$
$$ 2^{31} = 2 \mod 11$$
Now by using Fermat's little theorem $$ ({2^{11}})^{31} = 2^{11} \mod 31 $$ but $2^{11} = 2 \mod 31$ so I leave you to fill the details of showing $2^{341} = 2 \mod 341$.
Infinity_hunter
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Although this is the standard "Burton counterexample", we should note that it is not really a counterexample to Fermat's Little Theorem (see this for details). – Sayan Dutta Aug 29 '22 at 12:34