Prime number and Divisors












2














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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  • 4




    A number with an odd number of divisors must be square.
    – Lord Shark the Unknown
    1 hour ago
















2














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







share|cite|improve this question




















  • 4




    A number with an odd number of divisors must be square.
    – Lord Shark the Unknown
    1 hour ago














2












2








2


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






elementary-number-theory prime-numbers







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




elementary-number-theory prime-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 48 mins ago









Chinnapparaj R

5,2271826




5,2271826










asked 1 hour ago









Mohammad Mizanur Rahaman

135




135








  • 4




    A number with an odd number of divisors must be square.
    – Lord Shark the Unknown
    1 hour ago














  • 4




    A number with an odd number of divisors must be square.
    – Lord Shark the Unknown
    1 hour ago








4




4




A number with an odd number of divisors must be square.
– Lord Shark the Unknown
1 hour ago




A number with an odd number of divisors must be square.
– Lord Shark the Unknown
1 hour ago










1 Answer
1






active

oldest

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5














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
    54 mins ago






  • 1




    @MohammadMizanurRahaman: yes, $p=q=2$
    – Ross Millikan
    44 mins ago











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1 Answer
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5














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
    54 mins ago






  • 1




    @MohammadMizanurRahaman: yes, $p=q=2$
    – Ross Millikan
    44 mins ago
















5














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
    54 mins ago






  • 1




    @MohammadMizanurRahaman: yes, $p=q=2$
    – Ross Millikan
    44 mins ago














5












5








5






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 1 hour ago









Barycentric_Bash

38728




38728












  • The answer I have got is 2(maximum value of P). I think I am right ?
    – Mohammad Mizanur Rahaman
    54 mins ago






  • 1




    @MohammadMizanurRahaman: yes, $p=q=2$
    – Ross Millikan
    44 mins ago


















  • The answer I have got is 2(maximum value of P). I think I am right ?
    – Mohammad Mizanur Rahaman
    54 mins ago






  • 1




    @MohammadMizanurRahaman: yes, $p=q=2$
    – Ross Millikan
    44 mins ago
















The answer I have got is 2(maximum value of P). I think I am right ?
– Mohammad Mizanur Rahaman
54 mins ago




The answer I have got is 2(maximum value of P). I think I am right ?
– Mohammad Mizanur Rahaman
54 mins ago




1




1




@MohammadMizanurRahaman: yes, $p=q=2$
– Ross Millikan
44 mins ago




@MohammadMizanurRahaman: yes, $p=q=2$
– Ross Millikan
44 mins ago


















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