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What does ''$p$ of order $10\mod 11$'' mean ?

$p$ is a prime, what are then the possibilities for $p$ ?

user257
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  • Do you desire to know the definition of order, or do you already know that, but wish to characterize the elements that have order $10$ modulo $11?$ – Bill Dubuque Jul 19 '16 at 15:49
  • @BillDubuque I didn't know the definition but Ethan has already answered it, by fooling around I found that $2,6,7,8$ are the elements, thanks – user257 Jul 19 '16 at 16:38
  • @user257 I don't think the answer in your comment is right. See my edited answer. – Ethan Bolker Jul 19 '16 at 18:01

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If you calculate the powers of $p$ mod 11 the first power you find whose value is 1 (mod 11) is the order of $p$ mod 11.

I assume that your second question means "what can the order of $p$ be for the various primes $p$? If you try out several you should be able to guess the answer.

For example, for $p=3$ we have $3^2 = 9$, $3^3 = 27 \equiv 5 \pmod {11}$, $3^4 \equiv 3 \times 5 = 15 \equiv 4 \pmod{11}$ and $3^5 \equiv 3 \times 4 = 12 \equiv 1 \pmod{11}$ so the order of the prime $3$ is $5$. That means $5$ is one of the possible orders asked for.

If you work this out for $p=2$ you will find that its order is $10$.

Ethan Bolker
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  • Note that there are infinitely many primes in each nonzero congruence class $\mod 11$, by Dirichlet's theorem on primes in arithmetic progressions. – Robert Israel Jul 19 '16 at 15:42
  • @RobertIsrael True - but it's a stretch for the OP at this stage of his or her studies to be able to say that the "prime" hypothesis is irrelevant. – Ethan Bolker Jul 19 '16 at 16:05
  • @Ethan Bolker Yes $3^5\equiv 1\mod 10$ but then the order is $5$, so it is out of question I need only the elements (primes)$\le11$ of order $10$, and these are $2,6,7,8$ – user257 Jul 19 '16 at 19:18
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    @user257 You have correctly found the four elements of order 10 (mod 11). That may be the question you wanted to ask, but it's not the one you did ask - that one was about primes. (Typo In your comment: you mean mod 11, not mod 10.) – Ethan Bolker Jul 19 '16 at 19:59