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Let $b,c \in \mathbb Z$. Suppose that $11$ does not divide $b$ and that $b^3\equiv c^3\mod 11$.

$(i)$ Show that there exists $z \in \mathbb Z$ such that $bz\equiv 1\mod 11$

$(ii)$ Show that $b\equiv c\mod 11$.

I have done $(i)$ but I have no idea how to do $(ii)$. Can someone please help me ?

Bill Dubuque
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  • Just check the few possible classes modulo $11$. (Work in the field $\Bbb Z/11\Bbb Z=\Bbb F_{11}$, the field with $11$ elements.) Check that the function $x\to x^3$ is injective or surjective seen as a function $\Bbb F_{11}\to\Bbb F_{11}$. – dan_fulea Jan 02 '23 at 15:23
  • @dan_fulea we haven't learnt to do questions like that. i would need to use a modular arithmetic method multiplying the congruences and then simplifying. –  Jan 02 '23 at 15:29
  • Same as in your question yesterday by taking cube roots. – Bill Dubuque Jan 02 '23 at 15:32
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    If there are such constraints for a solution, please always mention them in the question. Modular arithmetic is working inside (the ring / the field of) integers taken modulo $11$, which is in notation $\Bbb Z/11\Bbb Z$. In "modular arithmetic" the first step is to pass modulo $11$, so make a list of the map $x\to x^3$ for all possible classes modulo $11$, which are $0$ (excluded for $b$, but...), $\pm1$, $\pm 2$, $\pm 3$, $\pm 4$, and $\pm 5$. So compute these six cubes modulo $11$, getting respectively $0$, $(\pm 1)^3=\pm1$, $(\pm2)^3=\pm 8=\mp3$, and so on. Are there any "collisions"? – dan_fulea Jan 02 '23 at 15:45
  • @dan_fulea No, brute force residue case checking should not be the first step. Exercises like this are designed to be solved more conceptually, not by brute force (which would fail miserably for larger moduli). – Bill Dubuque Jan 02 '23 at 16:16
  • Note that $3 \cdot 7 \equiv 1 \bmod 10$. Then use Fermat. – lhf Jan 02 '23 at 16:53
  • @BillDubuque So you want me to give the hint, that the inverse map is $x\to x^7$, which is the structural way, so we work in $\Bbb F^\times_{11}$ (or in some $\Bbb F^\times_p$ for some favorite big $p$), we know that each element has order dividing $(p-1)$, Fermat, and then invert the $3$ modulo $(p-1)$ - if possible. – dan_fulea Jan 02 '23 at 17:19
  • @dan No, since that method is already in the dupe I linked (and many other answers) – Bill Dubuque Jan 02 '23 at 17:22
  • @BillDubuque OK, Bill, it is a matter of taste, i was inside the link, which linked me to an other link, where i saw your answer. That second question was using some bigger modulu. Instead of senting me from one question full of comments which has a good idea in the question, to an other question with an answer presented in an unstructural manner (to fit the ad-hoc manner of the question). Instead, please tell me the idea, it is too hard to extract it from somewhere else, not sure where now. I asked explicitly if using the inverse $x\to x^7$ should be "the way" - structural and quick enough. – dan_fulea Jan 02 '23 at 17:53
  • @dan It's not clear what links you refer to. I refer to the Theorem in the linked dupe. – Bill Dubuque Jan 02 '23 at 17:57
  • @BillDubuque The cited theorem in the "dupe you linked" was not easy to find. It is the same (with many letters) with the fact that $3\times7=1$ modulo ten, as in the comment of lhf. Then i asked immediately after the comment of lhf if you want me to give this hint, that $x\to x^7$ is the inverse of $x\to x^3$. Yes is the answer, and we can still compare (didactically or in a pragmatic manner or in a way to have a quick solution without objections) the two ways to proceed. When you start a comment with a <<No, ... should not be the first step>> give an argument why not. I still like mine. – dan_fulea Jan 02 '23 at 19:16
  • @dan The Theorem is in the accepted answer in the linked dupe I cited, so I'm puzzled why it was not easy to find. If you think that answer is "the same as 3 x 7 = 1" then we have greatly divergent views on pedagogy, This is not the place to discuss such. – Bill Dubuque Jan 02 '23 at 19:25
  • @BillDubuque Just to fix the level of communication, since there is a difference. The "linked dupe" must be maybe the linked duplicated something, (this must be arguably a joke, although we were not on the foot of making jokes), usually answers are duplicated, not solutions to such answers, so i clicked in your laconic comment that "Theorem" link, moved instantly to the top of the question, to see the question, had to read, no connection to this question, no Theorem. Then i looked below to where i was coming from. The "Theorem" is equivalent to $x\to x^k$ is the inverse of $x\to x^{k'}$. – dan_fulea Jan 02 '23 at 19:51
  • @dan It's already there, since a $k$'th root function denotes an inverse of the $k$'th power function. My notion of "dupe" is more abstract than yours see abstract duplicates.Btw, if that's your recent downvote, constructive criticism is always appreciated. – Bill Dubuque Jan 02 '23 at 20:01
  • @BillDubuque No, not my downvote, i highly respect and appreciate your effort and work, here and at any other place. Good organized thoughts, valid solutions, good display. (It is also ok to close this question from the point of view of the site. My position was trying to give a chance those that make the first steps in number theory... An other story, it may be that i am on a wrong side above, since finally i made the difference going into the structures...) – dan_fulea Jan 03 '23 at 13:52

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