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Prove that for any positive integer $a$, $a^{561} \equiv a \pmod{561}$. (Hence, $561$ is a pseudoprime with respect to any base. Such a number is called a Carmichael number.)

This obviously works for $1$ but how do I find $2^{561}$ or any other number to the power of $561?$

Michael Rozenberg
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qs13
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  • I looked at link but it didn't really help. – qs13 Jun 15 '20 at 14:38
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    Factor $561$ and do this for each of the prime factors. – lulu Jun 15 '20 at 14:38
  • I factored 561, but what do you mean by "do this for each of the prime factors"? – qs13 Jun 15 '20 at 14:39
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    We have $561=3\times 11\times 17$. So, just show that $a^{561}\equiv 1 \pmod p$ for $p\in {3,11,17}$. To get you started: the case $p=3$ can be done in your head. And $11$ is only slightly harder. – lulu Jun 15 '20 at 14:41
  • All the cases are the same. With $n=561=p_1p_2p_3$ you just need to confirm that $(p_i-1),|,(n-1)$ – lulu Jun 20 '20 at 23:16

3 Answers3

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Fermat's theorem implies that $a^{(p-1)n+1} \equiv a \bmod p$ for all primes $p$, all $a$ and all $n$.

Since $561=3\cdot 11\cdot 17$, apply this to $(p,n)=(3,560/2)$, $(11,560/10)$, $(17,560/16)$.

lhf
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  • Is there something that I need to do after (p,n)=(3,560/2), (11,560/10), (17,560/16)? – qs13 Jun 15 '20 at 14:58
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    @132479, nothing, except make sure that those fractions are really integers. – lhf Jun 15 '20 at 14:59
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Since $$a^{561}-a=\left(a^{560}-1\right)a,$$ we see that $a^{561}-a$ is divisible by $a^3-a$, by $a^{17}-a$ and by $a^{11}-a$, which says that it's divisible by $3$, by $17$ and by $11$, which says that it's divisible by $3\cdot17\cdot11=561$.

Michael Rozenberg
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Composite numbers $n>1$, which satisfy $a^{n-1}\equiv 1\pmod{n}$ for all positive integers $a$ with $\gcd(a,n)=1$, are called $\color{red}{\text{Carmichael numbers}}.$

There is a necessary and sufficient criterion for a positive integer to be a Carmichael number known as the Korselt's criterion

$\color{red}{\text{Korselt's Criterion:}}$ A positive integer $n>1$ is a Carmichael number if and only if $(1)$ $n$ is square-free, $(2)$ for any prime divisor $p$ of $n$, $p-1\mid n-1$

Proof: Try yourself. An easy application of Chinese Remainder Theorem

You can verify, using this criterion, that $561$ is a Carmichael number. In fact, there are infinitely many of them.

ShBh
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