Prime number and Divisors [closed]

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-1














Let $p$ be a prime number such that $p^2+12$ has exactly $5$ divisors. What is the maximum value of $p$ ?



I came across this question in a Math Olympiad Competition and had no idea how to solve it










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closed as off-topic by TheSimpliFire, Lord_Farin, user21820, Did, RRL Dec 25 '18 at 14:50


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – TheSimpliFire, Lord_Farin, user21820, Did, RRL
If this question can be reworded to fit the rules in the help center, please edit the question.








  • 5




    A number with an odd number of divisors must be square.
    – Lord Shark the Unknown
    Dec 25 '18 at 6:02















-1














Let $p$ be a prime number such that $p^2+12$ has exactly $5$ divisors. What is the maximum value of $p$ ?



I came across this question in a Math Olympiad Competition and had no idea how to solve it










share|cite|improve this question















closed as off-topic by TheSimpliFire, Lord_Farin, user21820, Did, RRL Dec 25 '18 at 14:50


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – TheSimpliFire, Lord_Farin, user21820, Did, RRL
If this question can be reworded to fit the rules in the help center, please edit the question.








  • 5




    A number with an odd number of divisors must be square.
    – Lord Shark the Unknown
    Dec 25 '18 at 6:02













-1












-1








-1


1





Let $p$ be a prime number such that $p^2+12$ has exactly $5$ divisors. What is the maximum value of $p$ ?



I came across this question in a Math Olympiad Competition and had no idea how to solve it










share|cite|improve this question















Let $p$ be a prime number such that $p^2+12$ has exactly $5$ divisors. What is the maximum value of $p$ ?



I came across this question in a Math Olympiad Competition and had no idea how to solve it







elementary-number-theory prime-numbers






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edited Dec 25 '18 at 6:19









Chinnapparaj R

5,3051826




5,3051826










asked Dec 25 '18 at 6:00









Mohammad Mizanur Rahaman

105




105




closed as off-topic by TheSimpliFire, Lord_Farin, user21820, Did, RRL Dec 25 '18 at 14:50


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – TheSimpliFire, Lord_Farin, user21820, Did, RRL
If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by TheSimpliFire, Lord_Farin, user21820, Did, RRL Dec 25 '18 at 14:50


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – TheSimpliFire, Lord_Farin, user21820, Did, RRL
If this question can be reworded to fit the rules in the help center, please edit the question.







  • 5




    A number with an odd number of divisors must be square.
    – Lord Shark the Unknown
    Dec 25 '18 at 6:02












  • 5




    A number with an odd number of divisors must be square.
    – Lord Shark the Unknown
    Dec 25 '18 at 6:02







5




5




A number with an odd number of divisors must be square.
– Lord Shark the Unknown
Dec 25 '18 at 6:02




A number with an odd number of divisors must be square.
– Lord Shark the Unknown
Dec 25 '18 at 6:02










2 Answers
2






active

oldest

votes


















7














Notice that a number that has 5 divisors must be in the form of $q^4$ for some prime $q$. So we get:



$q^4=p^2+12 implies (q^2-p)(q^2+p)=12$



Then do some casework on it.






share|cite|improve this answer




















  • The answer I have got is 2(maximum value of P). I think I am right ?
    – Mohammad Mizanur Rahaman
    Dec 25 '18 at 6:13






  • 1




    @MohammadMizanurRahaman: yes, $p=q=2$
    – Ross Millikan
    Dec 25 '18 at 6:24


















0














Using @LordSharktheUnknown's hint that a number with an odd number of divisors must be square



we are then looking for cases where $p^2+12=a^2$ i.e. $a^2-p^2 = 12$



Since $7^2-6^2=13$, this tells you $a^2-p^2 gt 12$ for $a ge 7$ and $p lt a$



so here we must have $a lt 7$ and thus prime $p in 2,3,5$. Considering these:




  • $2^2+12=16=4^2$ so $p=2$ is indeed a possibility


  • $3^2+12=21$, not a square, so $p=3$ is not a possibility


  • $5^2+12=37$, not a square, so $p=5$ is not a possibility





share|cite|improve this answer



























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    7














    Notice that a number that has 5 divisors must be in the form of $q^4$ for some prime $q$. So we get:



    $q^4=p^2+12 implies (q^2-p)(q^2+p)=12$



    Then do some casework on it.






    share|cite|improve this answer




















    • The answer I have got is 2(maximum value of P). I think I am right ?
      – Mohammad Mizanur Rahaman
      Dec 25 '18 at 6:13






    • 1




      @MohammadMizanurRahaman: yes, $p=q=2$
      – Ross Millikan
      Dec 25 '18 at 6:24















    7














    Notice that a number that has 5 divisors must be in the form of $q^4$ for some prime $q$. So we get:



    $q^4=p^2+12 implies (q^2-p)(q^2+p)=12$



    Then do some casework on it.






    share|cite|improve this answer




















    • The answer I have got is 2(maximum value of P). I think I am right ?
      – Mohammad Mizanur Rahaman
      Dec 25 '18 at 6:13






    • 1




      @MohammadMizanurRahaman: yes, $p=q=2$
      – Ross Millikan
      Dec 25 '18 at 6:24













    7












    7








    7






    Notice that a number that has 5 divisors must be in the form of $q^4$ for some prime $q$. So we get:



    $q^4=p^2+12 implies (q^2-p)(q^2+p)=12$



    Then do some casework on it.






    share|cite|improve this answer












    Notice that a number that has 5 divisors must be in the form of $q^4$ for some prime $q$. So we get:



    $q^4=p^2+12 implies (q^2-p)(q^2+p)=12$



    Then do some casework on it.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 25 '18 at 6:05









    Barycentric_Bash

    41339




    41339











    • The answer I have got is 2(maximum value of P). I think I am right ?
      – Mohammad Mizanur Rahaman
      Dec 25 '18 at 6:13






    • 1




      @MohammadMizanurRahaman: yes, $p=q=2$
      – Ross Millikan
      Dec 25 '18 at 6:24
















    • The answer I have got is 2(maximum value of P). I think I am right ?
      – Mohammad Mizanur Rahaman
      Dec 25 '18 at 6:13






    • 1




      @MohammadMizanurRahaman: yes, $p=q=2$
      – Ross Millikan
      Dec 25 '18 at 6:24















    The answer I have got is 2(maximum value of P). I think I am right ?
    – Mohammad Mizanur Rahaman
    Dec 25 '18 at 6:13




    The answer I have got is 2(maximum value of P). I think I am right ?
    – Mohammad Mizanur Rahaman
    Dec 25 '18 at 6:13




    1




    1




    @MohammadMizanurRahaman: yes, $p=q=2$
    – Ross Millikan
    Dec 25 '18 at 6:24




    @MohammadMizanurRahaman: yes, $p=q=2$
    – Ross Millikan
    Dec 25 '18 at 6:24











    0














    Using @LordSharktheUnknown's hint that a number with an odd number of divisors must be square



    we are then looking for cases where $p^2+12=a^2$ i.e. $a^2-p^2 = 12$



    Since $7^2-6^2=13$, this tells you $a^2-p^2 gt 12$ for $a ge 7$ and $p lt a$



    so here we must have $a lt 7$ and thus prime $p in 2,3,5$. Considering these:




    • $2^2+12=16=4^2$ so $p=2$ is indeed a possibility


    • $3^2+12=21$, not a square, so $p=3$ is not a possibility


    • $5^2+12=37$, not a square, so $p=5$ is not a possibility





    share|cite|improve this answer

























      0














      Using @LordSharktheUnknown's hint that a number with an odd number of divisors must be square



      we are then looking for cases where $p^2+12=a^2$ i.e. $a^2-p^2 = 12$



      Since $7^2-6^2=13$, this tells you $a^2-p^2 gt 12$ for $a ge 7$ and $p lt a$



      so here we must have $a lt 7$ and thus prime $p in 2,3,5$. Considering these:




      • $2^2+12=16=4^2$ so $p=2$ is indeed a possibility


      • $3^2+12=21$, not a square, so $p=3$ is not a possibility


      • $5^2+12=37$, not a square, so $p=5$ is not a possibility





      share|cite|improve this answer























        0












        0








        0






        Using @LordSharktheUnknown's hint that a number with an odd number of divisors must be square



        we are then looking for cases where $p^2+12=a^2$ i.e. $a^2-p^2 = 12$



        Since $7^2-6^2=13$, this tells you $a^2-p^2 gt 12$ for $a ge 7$ and $p lt a$



        so here we must have $a lt 7$ and thus prime $p in 2,3,5$. Considering these:




        • $2^2+12=16=4^2$ so $p=2$ is indeed a possibility


        • $3^2+12=21$, not a square, so $p=3$ is not a possibility


        • $5^2+12=37$, not a square, so $p=5$ is not a possibility





        share|cite|improve this answer












        Using @LordSharktheUnknown's hint that a number with an odd number of divisors must be square



        we are then looking for cases where $p^2+12=a^2$ i.e. $a^2-p^2 = 12$



        Since $7^2-6^2=13$, this tells you $a^2-p^2 gt 12$ for $a ge 7$ and $p lt a$



        so here we must have $a lt 7$ and thus prime $p in 2,3,5$. Considering these:




        • $2^2+12=16=4^2$ so $p=2$ is indeed a possibility


        • $3^2+12=21$, not a square, so $p=3$ is not a possibility


        • $5^2+12=37$, not a square, so $p=5$ is not a possibility






        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 25 '18 at 9:12









        Henry

        98.4k475162




        98.4k475162












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