Prime number and Divisors [closed]

-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

edited Dec 25 '18 at 6:19

Chinnapparaj R

5,3341828

asked Dec 25 '18 at 6:00

Mohammad Mizanur Rahaman

155

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

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

add a comment |

-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

edited Dec 25 '18 at 6:19

Chinnapparaj R

5,3341828

asked Dec 25 '18 at 6:00

Mohammad Mizanur Rahaman

155

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

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

add a comment |

-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

edited Dec 25 '18 at 6:19

Chinnapparaj R

5,3341828

asked Dec 25 '18 at 6:00

Mohammad Mizanur Rahaman

155

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

edited Dec 25 '18 at 6:19

Chinnapparaj R

5,3341828

asked Dec 25 '18 at 6:00

Mohammad Mizanur Rahaman

155

edited Dec 25 '18 at 6:19

Chinnapparaj R

5,3341828

asked Dec 25 '18 at 6:00

Mohammad Mizanur Rahaman

155

edited Dec 25 '18 at 6:19

Chinnapparaj R

5,3341828

edited Dec 25 '18 at 6:19

Chinnapparaj R

5,3341828

edited Dec 25 '18 at 6:19

Chinnapparaj R

5,3341828

asked Dec 25 '18 at 6:00

Mohammad Mizanur Rahaman

155

asked Dec 25 '18 at 6:00

Mohammad Mizanur Rahaman

155

asked Dec 25 '18 at 6:00

Mohammad Mizanur Rahaman

155

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

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

add a comment |

5

$begingroup$
A number with an odd number of divisors must be square.
$endgroup$
– 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

add a comment |

2 Answers
2

active

oldest

votes

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.

answered Dec 25 '18 at 6:05

Barycentric_Bash

41339

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

1

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

add a comment |

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

answered Dec 25 '18 at 9:12

Henry

99.3k479165

add a comment |

2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

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.

answered Dec 25 '18 at 6:05

Barycentric_Bash

41339

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

1

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

add a comment |

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.

answered Dec 25 '18 at 6:05

Barycentric_Bash

41339

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

1

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

add a comment |

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.

answered Dec 25 '18 at 6:05

Barycentric_Bash

41339

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.

answered Dec 25 '18 at 6:05

Barycentric_Bash

41339

answered Dec 25 '18 at 6:05

Barycentric_Bash

41339

answered Dec 25 '18 at 6:05

Barycentric_Bash

41339

answered Dec 25 '18 at 6:05

Barycentric_Bash

41339

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

1

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

add a comment |

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

1

$begingroup$
@MohammadMizanurRahaman: yes, $p=q=2$
$endgroup$
– 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

@MohammadMizanurRahaman: yes, $p=q=2$

– Ross Millikan
Dec 25 '18 at 6:24

add a comment |

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

answered Dec 25 '18 at 9:12

Henry

99.3k479165

add a comment |

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

answered Dec 25 '18 at 9:12

Henry

99.3k479165

add a comment |

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

answered Dec 25 '18 at 9:12

Henry

99.3k479165

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

answered Dec 25 '18 at 9:12

Henry

99.3k479165

answered Dec 25 '18 at 9:12

Henry

99.3k479165

answered Dec 25 '18 at 9:12

Henry

99.3k479165

answered Dec 25 '18 at 9:12

Henry

99.3k479165

add a comment |

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