Divisors of a cyclotomic polynomial












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I am not entirely sure of the validity of this source (https://proofwiki.org/wiki/Prime_Divisors_of_Cyclotomic_Polynomials), but it claims that all prime divisors $d$ of $Phi_n(a)$ have $dequiv 1mod n$.



I am able to prove that $gcd(d-1, n)>1$, but I am not able to reconcile the above statement to myself. For example, for $Phi_{p^a}(a)$, are all prime factors still given by $dequiv 1mod p^a$. Or is it simply true that $dequiv 1mod p$?



Can someone offer a straightforward proof of this? A paper or official reference of this?






reference-request divisibility cyclotomic-polynomials







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    0












    $begingroup$


    I am not entirely sure of the validity of this source (https://proofwiki.org/wiki/Prime_Divisors_of_Cyclotomic_Polynomials), but it claims that all prime divisors $d$ of $Phi_n(a)$ have $dequiv 1mod n$.



    I am able to prove that $gcd(d-1, n)>1$, but I am not able to reconcile the above statement to myself. For example, for $Phi_{p^a}(a)$, are all prime factors still given by $dequiv 1mod p^a$. Or is it simply true that $dequiv 1mod p$?



    Can someone offer a straightforward proof of this? A paper or official reference of this?






    reference-request divisibility cyclotomic-polynomials







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      I am not entirely sure of the validity of this source (https://proofwiki.org/wiki/Prime_Divisors_of_Cyclotomic_Polynomials), but it claims that all prime divisors $d$ of $Phi_n(a)$ have $dequiv 1mod n$.



      I am able to prove that $gcd(d-1, n)>1$, but I am not able to reconcile the above statement to myself. For example, for $Phi_{p^a}(a)$, are all prime factors still given by $dequiv 1mod p^a$. Or is it simply true that $dequiv 1mod p$?



      Can someone offer a straightforward proof of this? A paper or official reference of this?






      reference-request divisibility cyclotomic-polynomials







      share|cite|improve this question









      $endgroup$




      I am not entirely sure of the validity of this source (https://proofwiki.org/wiki/Prime_Divisors_of_Cyclotomic_Polynomials), but it claims that all prime divisors $d$ of $Phi_n(a)$ have $dequiv 1mod n$.



      I am able to prove that $gcd(d-1, n)>1$, but I am not able to reconcile the above statement to myself. For example, for $Phi_{p^a}(a)$, are all prime factors still given by $dequiv 1mod p^a$. Or is it simply true that $dequiv 1mod p$?



      Can someone offer a straightforward proof of this? A paper or official reference of this?






      reference-request divisibility cyclotomic-polynomials




      reference-request divisibility cyclotomic-polynomials






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




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      asked Dec 13 '18 at 15:50









      Tejas RaoTejas Rao

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      30611






















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          I think your statement is wrong. For example, $Phi_3(x)=x^2+x+1$, $Phi_3(1)=3$ but $3equiv 0pmod 3$. However, we can prove the following statement (which is stated in your link).



          Proposition. Suppose that for some positive integer n and integer $a$ there prime number $p$ such that $p|Phi_n(a)$. Then, $p|n$ or $n|p-1$.



          Proof. Consider case when $n$ is not divisible by $p$. We need to prove that $p|n-1$. Let $s$ be a minimal positive integer such that $p|a^s-1$. Assume that $s<n$ (it's obvious that $sleq n$ because $p|a^n-1$). It's easy to see that $s|n$, so $n=ms$ for some positve integer $m$. Note that
          $$
          frac{a^{n}-1}{a^{s}-1}=prod_{d|n,~dnmid s}Phi_d(a),
          $$
          so $Phi_n(a)|frac{a^{n}-1}{a^{s}-1}$ (at this point we use the inequality $s<n$). Since $p|Phi_n(a)$ we obtain $p|frac{a^{n}-1}{a^{s}-1}$. On the other hand,
          $$
          frac{a^{n}-1}{a^{s}-1}=frac{a^{ms}-1}{a^{s}-1}=a^{(m-1)s}+a^{(m-2)s}+ldots+a^{s}+1equiv mpmod p
          $$
          because $a^sequiv 1pmod p$. Therefore, $p|m$, which impossible because $m|n$ and $pnmid n$. We get a contradiction, so our assumption was incorrect. Hence, $s=n$.
          Using Fermat's Little Theorem we obtain $p|a^{p-1}-1$. But $n$ is minimal $k$ such that $p|a^k-1$, so $n|p-1$ and proposition is proved.






          share|cite|improve this answer











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            $begingroup$

            I think your statement is wrong. For example, $Phi_3(x)=x^2+x+1$, $Phi_3(1)=3$ but $3equiv 0pmod 3$. However, we can prove the following statement (which is stated in your link).



            Proposition. Suppose that for some positive integer n and integer $a$ there prime number $p$ such that $p|Phi_n(a)$. Then, $p|n$ or $n|p-1$.



            Proof. Consider case when $n$ is not divisible by $p$. We need to prove that $p|n-1$. Let $s$ be a minimal positive integer such that $p|a^s-1$. Assume that $s<n$ (it's obvious that $sleq n$ because $p|a^n-1$). It's easy to see that $s|n$, so $n=ms$ for some positve integer $m$. Note that
            $$
            frac{a^{n}-1}{a^{s}-1}=prod_{d|n,~dnmid s}Phi_d(a),
            $$
            so $Phi_n(a)|frac{a^{n}-1}{a^{s}-1}$ (at this point we use the inequality $s<n$). Since $p|Phi_n(a)$ we obtain $p|frac{a^{n}-1}{a^{s}-1}$. On the other hand,
            $$
            frac{a^{n}-1}{a^{s}-1}=frac{a^{ms}-1}{a^{s}-1}=a^{(m-1)s}+a^{(m-2)s}+ldots+a^{s}+1equiv mpmod p
            $$
            because $a^sequiv 1pmod p$. Therefore, $p|m$, which impossible because $m|n$ and $pnmid n$. We get a contradiction, so our assumption was incorrect. Hence, $s=n$.
            Using Fermat's Little Theorem we obtain $p|a^{p-1}-1$. But $n$ is minimal $k$ such that $p|a^k-1$, so $n|p-1$ and proposition is proved.






            share|cite|improve this answer











            $endgroup$


















              1












              $begingroup$

              I think your statement is wrong. For example, $Phi_3(x)=x^2+x+1$, $Phi_3(1)=3$ but $3equiv 0pmod 3$. However, we can prove the following statement (which is stated in your link).



              Proposition. Suppose that for some positive integer n and integer $a$ there prime number $p$ such that $p|Phi_n(a)$. Then, $p|n$ or $n|p-1$.



              Proof. Consider case when $n$ is not divisible by $p$. We need to prove that $p|n-1$. Let $s$ be a minimal positive integer such that $p|a^s-1$. Assume that $s<n$ (it's obvious that $sleq n$ because $p|a^n-1$). It's easy to see that $s|n$, so $n=ms$ for some positve integer $m$. Note that
              $$
              frac{a^{n}-1}{a^{s}-1}=prod_{d|n,~dnmid s}Phi_d(a),
              $$
              so $Phi_n(a)|frac{a^{n}-1}{a^{s}-1}$ (at this point we use the inequality $s<n$). Since $p|Phi_n(a)$ we obtain $p|frac{a^{n}-1}{a^{s}-1}$. On the other hand,
              $$
              frac{a^{n}-1}{a^{s}-1}=frac{a^{ms}-1}{a^{s}-1}=a^{(m-1)s}+a^{(m-2)s}+ldots+a^{s}+1equiv mpmod p
              $$
              because $a^sequiv 1pmod p$. Therefore, $p|m$, which impossible because $m|n$ and $pnmid n$. We get a contradiction, so our assumption was incorrect. Hence, $s=n$.
              Using Fermat's Little Theorem we obtain $p|a^{p-1}-1$. But $n$ is minimal $k$ such that $p|a^k-1$, so $n|p-1$ and proposition is proved.






              share|cite|improve this answer











              $endgroup$
















                1












                1








                1





                $begingroup$

                I think your statement is wrong. For example, $Phi_3(x)=x^2+x+1$, $Phi_3(1)=3$ but $3equiv 0pmod 3$. However, we can prove the following statement (which is stated in your link).



                Proposition. Suppose that for some positive integer n and integer $a$ there prime number $p$ such that $p|Phi_n(a)$. Then, $p|n$ or $n|p-1$.



                Proof. Consider case when $n$ is not divisible by $p$. We need to prove that $p|n-1$. Let $s$ be a minimal positive integer such that $p|a^s-1$. Assume that $s<n$ (it's obvious that $sleq n$ because $p|a^n-1$). It's easy to see that $s|n$, so $n=ms$ for some positve integer $m$. Note that
                $$
                frac{a^{n}-1}{a^{s}-1}=prod_{d|n,~dnmid s}Phi_d(a),
                $$
                so $Phi_n(a)|frac{a^{n}-1}{a^{s}-1}$ (at this point we use the inequality $s<n$). Since $p|Phi_n(a)$ we obtain $p|frac{a^{n}-1}{a^{s}-1}$. On the other hand,
                $$
                frac{a^{n}-1}{a^{s}-1}=frac{a^{ms}-1}{a^{s}-1}=a^{(m-1)s}+a^{(m-2)s}+ldots+a^{s}+1equiv mpmod p
                $$
                because $a^sequiv 1pmod p$. Therefore, $p|m$, which impossible because $m|n$ and $pnmid n$. We get a contradiction, so our assumption was incorrect. Hence, $s=n$.
                Using Fermat's Little Theorem we obtain $p|a^{p-1}-1$. But $n$ is minimal $k$ such that $p|a^k-1$, so $n|p-1$ and proposition is proved.






                share|cite|improve this answer











                $endgroup$



                I think your statement is wrong. For example, $Phi_3(x)=x^2+x+1$, $Phi_3(1)=3$ but $3equiv 0pmod 3$. However, we can prove the following statement (which is stated in your link).



                Proposition. Suppose that for some positive integer n and integer $a$ there prime number $p$ such that $p|Phi_n(a)$. Then, $p|n$ or $n|p-1$.



                Proof. Consider case when $n$ is not divisible by $p$. We need to prove that $p|n-1$. Let $s$ be a minimal positive integer such that $p|a^s-1$. Assume that $s<n$ (it's obvious that $sleq n$ because $p|a^n-1$). It's easy to see that $s|n$, so $n=ms$ for some positve integer $m$. Note that
                $$
                frac{a^{n}-1}{a^{s}-1}=prod_{d|n,~dnmid s}Phi_d(a),
                $$
                so $Phi_n(a)|frac{a^{n}-1}{a^{s}-1}$ (at this point we use the inequality $s<n$). Since $p|Phi_n(a)$ we obtain $p|frac{a^{n}-1}{a^{s}-1}$. On the other hand,
                $$
                frac{a^{n}-1}{a^{s}-1}=frac{a^{ms}-1}{a^{s}-1}=a^{(m-1)s}+a^{(m-2)s}+ldots+a^{s}+1equiv mpmod p
                $$
                because $a^sequiv 1pmod p$. Therefore, $p|m$, which impossible because $m|n$ and $pnmid n$. We get a contradiction, so our assumption was incorrect. Hence, $s=n$.
                Using Fermat's Little Theorem we obtain $p|a^{p-1}-1$. But $n$ is minimal $k$ such that $p|a^k-1$, so $n|p-1$ and proposition is proved.







                share|cite|improve this answer














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                edited Jan 12 at 21:27

























                answered Jan 12 at 14:39









                richrowrichrow

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