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Let n=561 and let a be an integer. Suppose that a is coprime to n. Show that $a^{n-1}$ is congruent to 1 mod n.

user129221
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3 Answers3

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Hint: The prime factors of 561 are 3, 11 and 17. Notice that if $n=561$ and $p$ is one of its prime factors then $p - 1 $ divides $n - 1$ (just check the three cases). Finish the exercise by applying Fermat's little theorem and the Chinese remainder theorem.

Further reading: https://en.wikipedia.org/wiki/Carmichael_number

Yoni Rozenshein
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  • Why is it significant that p-1 divides n-1? I'm still really stuck. – user129221 Mar 11 '14 at 21:50
  • If $n-1 = k(p-1)$ then raising something to the $n-1$'th power is like raising it to the $p-1$'th power and then to another power: $a^{n-1} = \left(a^{p-1}\right)^k$. Now try fixing $p$ (one of 3, 11, and 17) and see where that gets you. – Yoni Rozenshein Mar 11 '14 at 21:53
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You need to prove $561 \mid a^{560} -1$. $560=3\cdot 11\cdot 17$ so you have to prove divisibility by 3, 11 and 17. Note that $\varphi(3), \varphi(11), \varphi(17) \mid 560$ what by Euler's theorem gives you what you need.

user101521
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Hint $\ $ Either apply Carmichael's generalization of Euler-Fermat, or proceed directly via

$$\rm A^{N_j}\equiv 1\ \ (mod\ M_j)\ \Rightarrow\ A^{lcm\ N_j}\equiv 1\ \ (mod\ lcm\ M_j)\ \ \ for\ \ \ \begin{cases} \,N = (2,10,16)\\ \rm M = (3,11,17)\end{cases}$$

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Bill Dubuque
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