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Let $n$ and $a>1$ be an integer and $n=q_1^{r_1}\cdots q_s^{r_s}$ is prime decomposition.

The book now I read(Gerald J.Janusz "Algebraic Number Fields") already prove following fact(Lemma 5.3).

Let $a,r$ be integers, each at least 2, and $q$ a prime integer. Then there exist a prime $p$ such that the multiplicative order of $a$ modulo $p$ is $q^r$.

So, the book says "For any $r\geq r_i$ and $r\geq 2$ there is a prime $p_i$ such that $a$ has order $q_i^{r_i}$"(maybe the book mistake $q_i^r$ for $q_i^{r_i}$)

Then the book says "As r increases, $p_i$ also increases and the order of $a$ mod $p_i$ is divisible by $q_i^{r_i}$". But I don't understand "As r increases, $p_i$ also increases". Would someone tell me?

Edit This is also by me An problem in elementary number theory used for proving Artin's lemma(class field theory)

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I didn’t read this book and I don’t know what the author meant, so I thought as follows. It is possible that for given $a$, $r$, and $q$ there are several primes $p$ such that multiplicative order $\deg_p a$ of $a$ modulo $p$ is $q^r$. For instance, for $a=13$, $r=2$, and $q=2$, $\deg_p a=q^r$ that is $\deg_p 13=4$ both for $p=5$ and $p=17$. If we have to choose one $p=p(r)$ among such primes, it is natural to put $p(r)$ the smallest prime $p$ such that $\deg_p a=q^r$. But even in this case it can be $p(r+1)<p(r)$. For instance, for $a=19$ and $q=2$, we have $p(2)=181$, but $p(3)=17$.

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You have $a,r_i$ and $q_i$ are fixed now there is a prime $p_i$ for each $i$ such that $a^{q_i^{r_i} }\cong 1 (\mod p_i )$ the book says if we consider $r\geq r_i$ then for any such $r$ there exists $p$ with $a^{q_i^r}\cong 1 (\mod p)$ and the statements follows if $r$ increases $p$ increases.

If you look at lemma 5.2 you will notice the book mean by $a$ have order $q^r$ mod $p$ is $$a^{q^r}\cong 1 (\mod p)$$ And $\textbf{r}$ is the least integer satisfying this not the least $q^r$. Now let $r'=r+s$ and suppose the order of $a$ mod $p$ is $q^r$ and mod $p'$ is $q^{r'}$ I want to show that $p'>p$ , $$a^{q^{r'}}=(a^{q^r})^{q^s}\cong 1 (\mod p')$$ keep in mind that $p'$ satisfying $$p'\mid a^{q^{r'}}-1 ,\; p'\nmid a^{q^{k}}-1: k<r'$$ And $$a^{q^{r'}}-1= (a^{q^{r}})^{q^s}-1=(a^{q^{r}}-1)P(a^{q^r})$$ I factored using $x^n-1=(x-1)(x^{n-1}+\cdots 1)$ in the last step where $P$ is a polynomial. So we have e $a^{q^{r}}-1\mid a^{q^{r'}}-1$ and we know that $p'\nmid a^{q^{r}}-1$ because $r<r'$ . I hope this helps you to figure why $p<p'$

I will write the two lemmas in case somebody else can help

5.2 lemma: Let $a$ and $r$ be integers $\geq 2$ and $q$ a prime integer. There exists a prime $p$ such that $a$ has order $q^r$ modulo $p$.

Note in the proof the book pick $p \mid X^{q-1} + X^{q-2} + \cdots + X+1 $ where $X=a^{q^{r-1}}$. Then proved that $r$ must be the least integer such that $a^{q^r}\cong 1 \mod p$ so it is not the normal multiplicative order.

5.3 Lemma: Let $n=\prod_{i=1}^{s} q_i^{r_i}$ be the prime factorization of $n$ as distinct primes. Let $a>1 $ be an integer. There exists infinitely many square free integers $$m=p_1p_2\cdots p_sp'_1\cdots p_s'$$ such that the order of $a$ modulo $m$ is divisible by $n$.

The book begins the proof as the op mentioned above.

More ideas

Note that: $a$ is relatively prime with $p$ and $p'$ since $a^{q^r} \equiv 1 (\mod p)$ and $a^{q^{r'}} \equiv 1 (\mod p')$

Now by Euler theorem we have $a^{p-1} \equiv 1 (\mod p) $ and $a^{p'-1}\equiv 1 (\mod p')$

But $r,r'$ are the least integers such that $a^{q^r}\equiv 1 (\mod p) , a^{q^{r'}} \equiv 1 (\mod p')$ hence we have $q^{r'} \mid p'-1 $ and $q^{r} \mid p-1 $.

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  • $\begingroup$ Then why if $r$ increases $p$ increases? $\endgroup$
    – user682141
    Sep 21, 2019 at 17:01
  • $\begingroup$ @user682141 I edited my answer $\endgroup$
    – IrbidMath
    Sep 28, 2019 at 12:10
  • $\begingroup$ Then why $p<p'$ at last? $\endgroup$
    – user682141
    Sep 29, 2019 at 3:11
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    $\begingroup$ I think it is normal multiplicative order because the order must be the form $q^i$ $\endgroup$
    – user682141
    Sep 29, 2019 at 3:13
  • $\begingroup$ Okay I see you are right because the order should divides $q^r$ so it is on the form $q^i$ for some $i$ $\endgroup$
    – IrbidMath
    Sep 29, 2019 at 16:44

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