Prime number and Divisors [closed]
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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
5,3051826
asked Dec 25 '18 at 6:00
Mohammad Mizanur Rahaman
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
add a comment |
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
5,3051826
asked Dec 25 '18 at 6:00
Mohammad Mizanur Rahaman
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
5
A number with an odd number of divisors must be square.
– Lord Shark the Unknown
Dec 25 '18 at 6:02
add a comment |
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
5,3051826
asked Dec 25 '18 at 6:00
Mohammad Mizanur Rahaman
105
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
edited Dec 25 '18 at 6:19
Chinnapparaj R
5,3051826
asked Dec 25 '18 at 6:00
Mohammad Mizanur Rahaman
105
edited Dec 25 '18 at 6:19
Chinnapparaj R
5,3051826
asked Dec 25 '18 at 6:00
Mohammad Mizanur Rahaman
105
edited Dec 25 '18 at 6:19
Chinnapparaj R
5,3051826
edited Dec 25 '18 at 6:19
Chinnapparaj R
5,3051826
edited Dec 25 '18 at 6:19
Chinnapparaj R
5,3051826
5,3051826
asked Dec 25 '18 at 6:00
Mohammad Mizanur Rahaman
105
asked Dec 25 '18 at 6:00
Mohammad Mizanur Rahaman
105
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
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
5
A number with an odd number of divisors must be square.
– Lord Shark the Unknown
Dec 25 '18 at 6:02
add a comment |
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
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
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
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
98.4k475162
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
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
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
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
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
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
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
add a comment |
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
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
98.4k475162
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
98.4k475162
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
98.4k475162
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
98.4k475162
answered Dec 25 '18 at 9:12
Henry
98.4k475162
answered Dec 25 '18 at 9:12
Henry
98.4k475162
answered Dec 25 '18 at 9:12
Henry
98.4k475162
98.4k475162
add a comment |
add a comment |
5
A number with an odd number of divisors must be square.
– Lord Shark the Unknown
Dec 25 '18 at 6:02