If $p geq 11$ is a prime then either $p^3-1$ or $p^3+1$ is divisible by $14$












0












$begingroup$



True or False:



If $p$ is a prime greater than or equal to $ 11$, then either $p^3-1$ or $p^3+1$ is divisible by $14$





My try



The statement is true. In order to prove this, we prove either $$p^3equiv1(;text{mod};14;);;text{or};;p^3equiv -1(;text{mod};14;)$$ Since $gcd(14,p)=1$, by Euler's theorem , $$p^{large phi(14)}equiv1(;text{mod};14;)$$ That is $$p^6equiv1(;text{mod};14;)$$ From the above, can I conclude the desired result ? Any help ?






elementary-number-theory divisibility







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  • $begingroup$
    $p^6equiv 1 implies p^6 - 1equiv 0implies (p^3 + 1)(p^3-1)equiv 0$.
    $endgroup$
    – fleablood
    Dec 23 '18 at 23:07
















0












$begingroup$



True or False:



If $p$ is a prime greater than or equal to $ 11$, then either $p^3-1$ or $p^3+1$ is divisible by $14$





My try



The statement is true. In order to prove this, we prove either $$p^3equiv1(;text{mod};14;);;text{or};;p^3equiv -1(;text{mod};14;)$$ Since $gcd(14,p)=1$, by Euler's theorem , $$p^{large phi(14)}equiv1(;text{mod};14;)$$ That is $$p^6equiv1(;text{mod};14;)$$ From the above, can I conclude the desired result ? Any help ?






elementary-number-theory divisibility







share|cite|improve this question









$endgroup$












  • $begingroup$
    $p^6equiv 1 implies p^6 - 1equiv 0implies (p^3 + 1)(p^3-1)equiv 0$.
    $endgroup$
    – fleablood
    Dec 23 '18 at 23:07














0












0








0





$begingroup$



True or False:



If $p$ is a prime greater than or equal to $ 11$, then either $p^3-1$ or $p^3+1$ is divisible by $14$





My try



The statement is true. In order to prove this, we prove either $$p^3equiv1(;text{mod};14;);;text{or};;p^3equiv -1(;text{mod};14;)$$ Since $gcd(14,p)=1$, by Euler's theorem , $$p^{large phi(14)}equiv1(;text{mod};14;)$$ That is $$p^6equiv1(;text{mod};14;)$$ From the above, can I conclude the desired result ? Any help ?






elementary-number-theory divisibility







share|cite|improve this question









$endgroup$





True or False:



If $p$ is a prime greater than or equal to $ 11$, then either $p^3-1$ or $p^3+1$ is divisible by $14$





My try



The statement is true. In order to prove this, we prove either $$p^3equiv1(;text{mod};14;);;text{or};;p^3equiv -1(;text{mod};14;)$$ Since $gcd(14,p)=1$, by Euler's theorem , $$p^{large phi(14)}equiv1(;text{mod};14;)$$ That is $$p^6equiv1(;text{mod};14;)$$ From the above, can I conclude the desired result ? Any help ?






elementary-number-theory divisibility




elementary-number-theory divisibility






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asked Dec 22 '18 at 10:18









Chinnapparaj RChinnapparaj R

6,54221029




6,54221029












  • $begingroup$
    $p^6equiv 1 implies p^6 - 1equiv 0implies (p^3 + 1)(p^3-1)equiv 0$.
    $endgroup$
    – fleablood
    Dec 23 '18 at 23:07


















  • $begingroup$
    $p^6equiv 1 implies p^6 - 1equiv 0implies (p^3 + 1)(p^3-1)equiv 0$.
    $endgroup$
    – fleablood
    Dec 23 '18 at 23:07
















$begingroup$
$p^6equiv 1 implies p^6 - 1equiv 0implies (p^3 + 1)(p^3-1)equiv 0$.
$endgroup$
– fleablood
Dec 23 '18 at 23:07




$begingroup$
$p^6equiv 1 implies p^6 - 1equiv 0implies (p^3 + 1)(p^3-1)equiv 0$.
$endgroup$
– fleablood
Dec 23 '18 at 23:07










3 Answers
3






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5












$begingroup$

Consider that for $pnotequiv0pmod7$
$$
left(p^3-1right)left(p^3+1right)=p^6-1equiv0pmod7
$$
So either $7mid p^3-1$ or $7mid p^3+1$. Since $p$ is odd, $2mid p^3-1$ and $2mid p^3+1$. Thus,
$$
14mid p^3-1quadtext{or}quad14mid p^3+1
$$






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Hint: $p^6equiv 1 mod 14$ means that $p^6-1 = (p^3-1)(p^3+1)$ is divisible by $14$. We have $2mid p^3+1$ and $2mid p^3-1$, since the numbers $p^3pm 1$ are even. Moreover, $7$ is prime and so $7mid p^3+1$ or $7mid p^3-1$. Thus $14$ divides $p^3+1$ or $p^3-1$.






    share|cite|improve this answer











    $endgroup$





















      0












      $begingroup$

      Since the other solutions helped you from what you had, I have another "bruteforce" solution:



      If $p > 11$ is prime, $2not mid p land 7not mid p implies bar{p} in {bar{1}, bar{3}, bar{5}, bar{-5}, bar{-3}, bar{-1}}$. $1^3 equiv 1, 5^3 equiv -1, 3^3 equiv -1$ and we're done.






      share|cite|improve this answer









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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

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        active

        oldest

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        5












        $begingroup$

        Consider that for $pnotequiv0pmod7$
        $$
        left(p^3-1right)left(p^3+1right)=p^6-1equiv0pmod7
        $$
        So either $7mid p^3-1$ or $7mid p^3+1$. Since $p$ is odd, $2mid p^3-1$ and $2mid p^3+1$. Thus,
        $$
        14mid p^3-1quadtext{or}quad14mid p^3+1
        $$






        share|cite|improve this answer









        $endgroup$


















          5












          $begingroup$

          Consider that for $pnotequiv0pmod7$
          $$
          left(p^3-1right)left(p^3+1right)=p^6-1equiv0pmod7
          $$
          So either $7mid p^3-1$ or $7mid p^3+1$. Since $p$ is odd, $2mid p^3-1$ and $2mid p^3+1$. Thus,
          $$
          14mid p^3-1quadtext{or}quad14mid p^3+1
          $$






          share|cite|improve this answer









          $endgroup$
















            5












            5








            5





            $begingroup$

            Consider that for $pnotequiv0pmod7$
            $$
            left(p^3-1right)left(p^3+1right)=p^6-1equiv0pmod7
            $$
            So either $7mid p^3-1$ or $7mid p^3+1$. Since $p$ is odd, $2mid p^3-1$ and $2mid p^3+1$. Thus,
            $$
            14mid p^3-1quadtext{or}quad14mid p^3+1
            $$






            share|cite|improve this answer









            $endgroup$



            Consider that for $pnotequiv0pmod7$
            $$
            left(p^3-1right)left(p^3+1right)=p^6-1equiv0pmod7
            $$
            So either $7mid p^3-1$ or $7mid p^3+1$. Since $p$ is odd, $2mid p^3-1$ and $2mid p^3+1$. Thus,
            $$
            14mid p^3-1quadtext{or}quad14mid p^3+1
            $$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 22 '18 at 10:25









            robjohn♦robjohn

            271k27316643




            271k27316643























                1












                $begingroup$

                Hint: $p^6equiv 1 mod 14$ means that $p^6-1 = (p^3-1)(p^3+1)$ is divisible by $14$. We have $2mid p^3+1$ and $2mid p^3-1$, since the numbers $p^3pm 1$ are even. Moreover, $7$ is prime and so $7mid p^3+1$ or $7mid p^3-1$. Thus $14$ divides $p^3+1$ or $p^3-1$.






                share|cite|improve this answer











                $endgroup$


















                  1












                  $begingroup$

                  Hint: $p^6equiv 1 mod 14$ means that $p^6-1 = (p^3-1)(p^3+1)$ is divisible by $14$. We have $2mid p^3+1$ and $2mid p^3-1$, since the numbers $p^3pm 1$ are even. Moreover, $7$ is prime and so $7mid p^3+1$ or $7mid p^3-1$. Thus $14$ divides $p^3+1$ or $p^3-1$.






                  share|cite|improve this answer











                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    Hint: $p^6equiv 1 mod 14$ means that $p^6-1 = (p^3-1)(p^3+1)$ is divisible by $14$. We have $2mid p^3+1$ and $2mid p^3-1$, since the numbers $p^3pm 1$ are even. Moreover, $7$ is prime and so $7mid p^3+1$ or $7mid p^3-1$. Thus $14$ divides $p^3+1$ or $p^3-1$.






                    share|cite|improve this answer











                    $endgroup$



                    Hint: $p^6equiv 1 mod 14$ means that $p^6-1 = (p^3-1)(p^3+1)$ is divisible by $14$. We have $2mid p^3+1$ and $2mid p^3-1$, since the numbers $p^3pm 1$ are even. Moreover, $7$ is prime and so $7mid p^3+1$ or $7mid p^3-1$. Thus $14$ divides $p^3+1$ or $p^3-1$.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Dec 24 '18 at 12:39

























                    answered Dec 22 '18 at 10:23









                    WuestenfuxWuestenfux

                    5,5581513




                    5,5581513























                        0












                        $begingroup$

                        Since the other solutions helped you from what you had, I have another "bruteforce" solution:



                        If $p > 11$ is prime, $2not mid p land 7not mid p implies bar{p} in {bar{1}, bar{3}, bar{5}, bar{-5}, bar{-3}, bar{-1}}$. $1^3 equiv 1, 5^3 equiv -1, 3^3 equiv -1$ and we're done.






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          Since the other solutions helped you from what you had, I have another "bruteforce" solution:



                          If $p > 11$ is prime, $2not mid p land 7not mid p implies bar{p} in {bar{1}, bar{3}, bar{5}, bar{-5}, bar{-3}, bar{-1}}$. $1^3 equiv 1, 5^3 equiv -1, 3^3 equiv -1$ and we're done.






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            Since the other solutions helped you from what you had, I have another "bruteforce" solution:



                            If $p > 11$ is prime, $2not mid p land 7not mid p implies bar{p} in {bar{1}, bar{3}, bar{5}, bar{-5}, bar{-3}, bar{-1}}$. $1^3 equiv 1, 5^3 equiv -1, 3^3 equiv -1$ and we're done.






                            share|cite|improve this answer









                            $endgroup$



                            Since the other solutions helped you from what you had, I have another "bruteforce" solution:



                            If $p > 11$ is prime, $2not mid p land 7not mid p implies bar{p} in {bar{1}, bar{3}, bar{5}, bar{-5}, bar{-3}, bar{-1}}$. $1^3 equiv 1, 5^3 equiv -1, 3^3 equiv -1$ and we're done.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 23 '18 at 22:02









                            Lucas HenriqueLucas Henrique

                            1,031414




                            1,031414






























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