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Find all $n \in{N}$ satisfying $2 \leq n \leq 30$ and $ord_n (x) < \phi(n), \;\forall x \in{Z}_n^*$.

${Z}_n^* = \{x\in{Z}_n\mid(x, n)= 1\} $

${Z}_n$ = set of all residue classes $\bmod n$

My work :

$ord_n (x) < \phi(n)$, so $x$ is not primitive root.

By Primitive root theorem, $n$ is in the form $p^k, 2p^k, 2, 4$ where $p\in$ odd prime.

For $n = 2$, we have $x^2 \equiv 1 (\bmod p)$ and $\phi(p) \geq 2$ for $p \geq 3 $.

For $n = 4$,

if $p \geq 5 $, we have $x^4 \equiv 1 (\bmod p)$ and $x^2 \equiv -1 (\bmod p)$ so $\phi(p) \geq 4$.

if $p = 3 $, this contradicts $x^2 \equiv -1 (\bmod p)$.

Is my work correct ?

Please suggest how to prove in case of $p^k$ and $2p^k$.

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    When you wrote: "By Primitive root theorem, $x$ is in the form $p^k, 2p^k, 2, 4$ where $p\in$ odd prime", don't you mean $n$, not $x$? – 2017-02-13
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    @quasi, edited. Thank you. – 2017-02-13

1 Answers 1

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I think you're not interpreting the condition

$$\forall x \in{Z}_n^*$$

correctly.

If there exists a primitive root mod $n$, the statement

$$ord_n (x) < \phi(n), \;\forall x \in{Z}_n^*$$

is false.

The statement is true if and only if $n$ is such that there does not exist a primitive root mod $n$. Thus, to answer the question, all you need to do is quote the primitive root theorem (as you did), and then list those values of $n$, where $2 \le n \le 30$, such that there does not exist a primitive root mod $n$.

Easy-peasy!

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    For each $n$, there may be many values of $x$ that $ (x,n) = 1$. Please clarify. – 2017-02-13
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    For an ineger $n > 1$, number of positive integer $x$(x,n)=1$ is $\phi(n)$. But that's not what the question is asking. It's not asking "how many values of x ...". It's asking "which values of $n$, $2 \le n \le 30$ are such that for _all_ $x \in Z_n^*$, ...". – 2017-02-13
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    $n = 2, 4, 3, 9, 27, 5, 25, 7, 11, 13, 17, 19, 23, 29, 6, 18, 10, 14, 22, 26$, right ? Could you please show me how to prove that $n = p^k$ and $2p^k$? – 2017-02-13
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    No, the answer is all values of $n$ which are _not_ one of $2,4,p^k,2p^k$. Thus, the answer to the given question is $$n = 8,12,15,16,20,21,24,28,30$$ – 2017-02-13
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    I'm careless. Thank you. For this problem, can we use the theorem without prove ? Is the prove for $p^k, 2p^k complicated ? – 2017-02-13
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    Yes, a little complicated. It can be proved using Elementary Number Theory, but the proofs which best explains the "why" of the result require some knowledge of Group Theory. – 2017-02-13
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    So for the assigned problem, just state the Primitive Root Theorem and use it to answer the given question. – 2017-02-13
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    However, if you're still curious, almost any textbook on Elementary Number Theory will have a proof of the Primitive Root Theorem. – 2017-02-13
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    Thank you for your helpful suggestion. – 2017-02-13