I'm asked to show that $a^{61} \equiv a\ (mod\ 1001)$ for every $a \in \mathbb{N}$. I've tried to tackle this using Fermat's Little Theorem and Euler's Theorem, but I can't even get started. My main problem seems to be the "for every $a \in \mathbb{N}$" part, because if it restricted $gcd(a, 1001) = 1$ I would be able to factor out 1001 and try to show that it's factors divide $a^{61} - a$. I'm kinda lost here. Any help is appreciated.
Show that $a^{61} \equiv a\ (mod\ 1001)$ for every $a \in \mathbb{N}$
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elementary-number-theory
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0Note that Fermat's little theorem, while usually stated in terms of $a^{p-1}\equiv 1$ and with the condition $\gcd(a,p)=1$, may just as well be stated as $a^p\equiv a$ _without_ the coprime condition. I think that will make it easier to see why that condition is not needed here either. – 2017-01-25
3 Answers
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Hint $\ $ Apply the following simple generalization of the little Fermat-Euler theorem. Since $\rm\:n = 1001 = \color{#C00}7\cdot\color{#0A0}{11}\cdot\color{brown}{13},\:$ is squarefree, it suffices to check $\rm\:\color{#C00}6,\color{#0A0}{10},\color{brown}{12}\:|\:61\!-\!1 = e\!-\!1.$
Theorem $\ $ For natural numbers $\rm\:a,e,n\:$ with $\rm\:e,n>1$
$\qquad\rm n\:|\:a^{\large e}-a\:$ for all $\rm\:a\:\iff n\:$ is squarefree, and prime $\rm\:p\:|\:n\:\Rightarrow\: p\!-\!1\:|\:e\!-\!1$
Proof $\ (\Leftarrow)\ \ $ Since a squarefree natural divides another iff all its
prime factors do, we need only show $\rm\:p\:|\:a^{\large e}\!-\!a\:$ for each prime $\rm\:p\:|\:n,\:$ or,
that $\rm\:a \not\equiv 0\:\Rightarrow\: a^{\large e-1} \equiv 1\pmod p,\:$ which, since $\rm\:p\!-\!1\:|\:e\!-\!1,\:$ follows
from $\rm\:a \not\equiv 0\:$ $\Rightarrow$ $\rm\: a^{\large p-1} \equiv 1 \pmod p,\:$ by little Fermat.
$(\Rightarrow)\ \ $ See this answer (not required here).
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Which version of Fermat's Little Theorem are you using? If you use
Theorem. Let $p$ be prime. Then $a^p\equiv a\pmod p$ for every integer $a$
you should find the problem works out quite simply. (Use a factorisation of $1001$, as you suggested.)
You can regard the above result as an alternative statement of Fermat's Little Theorem, or as a corollary of Fermat's Little Theorem.
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The smallest positive integer greater than $k$ such that $a^k \equiv 1 \bmod n$ for all $a$ coprime to $n$ is given by the charmicael lambda function. It is easy to caclulate it, as explained in this wikipedia page.
In the case in which $n$ is square free, $\lambda(n)+1$ is the smallest number such that $a^{\lambda(n)}\equiv a \bmod n$ for all $a$. We have that $\lambda(10001)+1=61$