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Can someone check my answers please. I am not confident on the first part but I think my second part is right. I think for the first part there is a better method to use.

Let $a \in \mathbb Z$. We have $a^3 ≡ 1$ mod $11$.

$(i)$ Show that $11$ does not divide $a$ and deduce that $a^{10} ≡ 1$ mod $11$.

Assume for sake of a contradiction that $11 | a$. Then we have $a ≡ 0$ mod $11$. So $a^3 ≡ 0$ mod $11$. By the symmetric property of $\mathbb Z_{11}$ since it is an equivalence relation, $0 ≡ a$ mod $11$. Then by the transitive property of $Z_{11}$, since $0 ≡ a^3$ mod $11$ and $a^3 ≡ 1$ mod $11$, then $0 ≡ 1$ mod $11$, which is clearly false. So we have a contradiction, meaning that $11$ does not divide $a$.

Now, using Fermat's Little Theorem (am I allowed to just state this), since $11$ is prime and $11$ does not divide $a$, we have $a^{10} ≡ 1$ mod $11$. $\square$

$(ii)$ Show that $a^{21} ≡ 1$ mod $11$ and deduce that $a ≡ 1$ mod $11$.

Since $a^3 ≡ 1$ mod $11$, we have $a^{21} ≡ 1$ mod $11$. We also have $a^{10} ≡ 1$ mod $11$, so we can see that $a^{20} ≡ 1$ mod $11$, and so $a^{21} ≡ a$ mod $11$. So again using the fact that $\mathbb Z_{11}$ is an equivalence relation, we have $a ≡ a^{21}$ mod $11$ and $a^{21} ≡ 1$ mod $11$, so by transitive property, $a ≡ 1$ mod $11$ as required. $\square$

Bill Dubuque
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    For the first part, it's a bit quicker to say that if $11\mid a$, then it also divides $a^3$, which is already the contradiction we wanted – Fix Jan 01 '23 at 14:19
  • @Fix Thanks - so basically you took the contrapositive of that statement ? And, does $(ii)$ look ok ? –  Jan 01 '23 at 14:28
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    To me it's more obvious to simply make a table of values and deduce that $a^3 \equiv 1 \iff a \equiv 1 \pmod{11}$. Though this does seem too brute force-y. – Hersh Jan 01 '23 at 14:29
  • @Hersh Actually that does make sense. It would have been easier to do that than thinking of my method and more simpler. Thanks for noting this –  Jan 01 '23 at 14:32
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    @NikitaMazepin your reasoning looks good, just be careful about the wording: $\mathbb{Z}_{11}$ is a ring and the $\equiv$ relation is the "=" in that ring, so multiplication and addition/subtraction works perfectly fine.$$$$ Also for deducing $a\equiv 1$ we could observe that $a^3\equiv a^{10}\equiv 1$, so the order of a divides both 10 and 3, meaning it can only be one – Fix Jan 01 '23 at 14:33
  • @Fix Okay thanks for your help : ) –  Jan 01 '23 at 14:35
  • By the linked dupe $\bmod 11!:\ a^3\equiv 1!\iff! \overbrace{(a^3)^{\color{#c00}7}\equiv 1^{\color{#c00}7}}^{\textstyle \ \ \ a\equiv 1},$ by taking $\rm\color{#0a0}{cube\ roots}$, by raising to power $,\color{#0a0}{1/3}\equiv 21/3\equiv 7\bmod{10}\ \ $ – Bill Dubuque Jan 01 '23 at 14:50
  • Your argument is correct. Written more concisely it is $$\bmod 11!:\ \color{#0a0}{a^3\equiv 1}\Rightarrow \color{#c00}{a\not\equiv 0}\Rightarrow \color{#0a0}1^7\equiv(\color{#0a0}{a^3})^7\equiv a(\color{#c00}{a^{10}})^2\overset{\color{#c00}{\rm F\ell T}}\equiv a(\color{#c00}1)^2\equiv a\qquad$$ Compare to the proof of the $k$'th root method in the dupe. Or: $,\color{#c00}{1\equiv a^{10}}\equiv a(\color{#0a0}{a^3})^3\equiv a(\color{#0a0}1)^3\equiv a.\ \ $ – Bill Dubuque Jan 01 '23 at 15:08
  • Generally $,a^j\equiv 1\equiv a^k\Rightarrow a^{\gcd(j,k)}\equiv 1,$ using $,\gcd(j,k) = mj+nk,$ by Bezout, and the above methods amount to different ways of computing the gcd. If 'order' is familar and if $,j,k,$ are coprime then the order of $a$ divides coprimes $,j,k,$ so the order must be $1,,$ i.e. $,a^1\equiv 1\ \ $ – Bill Dubuque Jan 01 '23 at 15:08
  • Once all is clear please delete the question since we already have far too many posts showing these methods. – Bill Dubuque Jan 01 '23 at 15:13
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    I recommend that you ignore the first comment. Your method using congruences is better than reverting back to manipulation of divisibility relations. One will quickly get lost in complexity if one always works with divisibility relations instead of congruences - which allow is to use or well-honed manipulations of equations (vs. far less intuitive (divisibility) relations). Ditto for the 3rd comment (the point of such exercises is to develop intuition for congruences (and group theory) - not brute-force (residue) case testing - which quickly becomes infeasible for larger problems). – Bill Dubuque Jan 01 '23 at 15:20
  • Please note that for future questions, for a solution-verification question to be on topic you must specify precisely which step in the proof you question, and why so. This site is not meant to be an open-ended proof checking machine. – Bill Dubuque Jan 01 '23 at 15:23
  • @BillDubuque So my proof by contradiction is okay ? –  Jan 01 '23 at 15:54
  • Yes. Normally at this level one doesn't justify every invocation of equivalence relation properties of congruences - just as in manipulation of ordinary equations (doing so obscures the heart of the matter). – Bill Dubuque Jan 01 '23 at 17:09

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