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$a \in \mathbb{Z}$ , $a$ and 11 are coprime numbers. So due to $Fermat$'s Theorem $a^{10} \equiv 1 \pmod {11} $. So I argued with my math teacher about the solutions of $a^n \equiv 1\pmod {11}$. He said and insisted that the solutions are $\mathbb{S} = \{10k \; \backslash \; k \in \mathbb{N} \}$. I disagreed and said that there are more cases to study :

  • If $a^{1} \equiv 1 \pmod {11}$: Then $\mathbb{S} = \mathbb{N}$
  • If $a^{2} \equiv 1 \pmod {11}$ and $a^{1} \not\equiv 1 \pmod {11}$: Then $\mathbb{S} = \{2k \; \backslash \; k \in \mathbb{N} \}$
  • If $a^{5} \equiv 1 \pmod {11}$ and $a^{1} \not\equiv 1 \pmod {11}$: Then $\mathbb{S} = \{5k \; \backslash \; k \in \mathbb{N} \}$
  • If $a^{10} \equiv 1 \pmod {11}$ and $a^{1} \not\equiv 1 \pmod {11}$ and $a^{2} \not\equiv 1 \pmod {11}$ and $a^{5} \not\equiv 1 \pmod {11}$: Then $\mathbb{S} = \{10k \; \backslash \; k \in \mathbb{N} \}$

So basically the point is that you should study the cases of all the positive divisors of 10 . Of course the teacher raged because he was achamed of me saying that , and I wanted to ask to see what is the correct answer . I'm all ears for your answers .

(EDIT : $a$ is a fixed number in $\mathbb{Z}$)

Put Me
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    It's a problem with quantifiers: if you're looking for a ‘universal’ $n$ (i.e. $a^n\equiv 1\mod 11$ for all $n$) then the solution is what your teacher said (and even the ideal $10\mathbf Z$). If you're looking for an $n$ depending on $a$, then you have to study cases . – Bernard Jun 04 '17 at 14:15
  • @Bernard , I guess you mean a universal $a$ . $a$ is a fixed number in $\mathbb{Z}$ – Put Me Jun 04 '17 at 14:49
  • No, I meant what I wrote. An $n$ suitable for all $a$ (not divisible by $11$). B.t.w., there was a typo in my comment: I meant‘$a^n\equiv 1\mod 11$ for all $a$’ of course. – Bernard Jun 04 '17 at 14:56
  • As i said, it's not for "all $a$" , it's for a fixed $a$ in $\mathbb{Z}$ – Put Me Jun 04 '17 at 15:14
  • If $a$ is fixed, it's the cases study which is correct. – Bernard Jun 04 '17 at 15:47

1 Answers1

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So there are two different questions. Find $n$ such that

  1. For all $a$ coprime with $11$, $a^n\equiv1\pmod{11}$.
  2. For a given $a$ coprime with $11$, $a^n \equiv 1\pmod {11}$.

Worth checking with the teacher which question they meant.

peterwhy
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  • It's the second question . $a$ is a fixed number in $\mathbb{Z}$ – Put Me Jun 04 '17 at 14:43
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    @PutMe: according to you! This looks like a simple misunderstanding to me: the teacher was probably answering the first question. – TonyK Jun 04 '17 at 14:54
  • @TonyK , it wasn't a misunderstanding actually , as this was a question in a math test and in the exercise beginning it was sais that $a$ is a fixed number in $\mathbb{Z}$ – Put Me Jun 04 '17 at 15:13