Prove that there exists only one prime number of the form $p^2−1$ where $p≥2$ is an integer
$begingroup$
By factoring $p^2 − 1$, we have $(p + 1)(p - 1)$.
I know that $p = 2$ which gives $3$ is the only solution.
However, how do I prove that $p = 2$ is the only integer which gives a prime?
elementary-number-theory discrete-mathematics proof-verification proof-writing
edited Jul 1 '18 at 20:09
Robert Soupe
11.5k21951
asked Dec 3 '15 at 1:44
Noah DengNoah Deng
8918
$endgroup$
add a comment |
$begingroup$
By factoring $p^2 − 1$, we have $(p + 1)(p - 1)$.
I know that $p = 2$ which gives $3$ is the only solution.
However, how do I prove that $p = 2$ is the only integer which gives a prime?
elementary-number-theory discrete-mathematics proof-verification proof-writing
edited Jul 1 '18 at 20:09
Robert Soupe
11.5k21951
asked Dec 3 '15 at 1:44
Noah DengNoah Deng
8918
$endgroup$
$begingroup$
Because of the $p-1$ factor...
$endgroup$
– abiessu
Dec 3 '15 at 1:45
$begingroup$
for any number>$2$, $p-1$ will give you a factor > $1$
$endgroup$
– Kushal Bhuyan
Dec 3 '15 at 1:46
$begingroup$
any other solution can't be prime because it can be decomposed as a product of two numbers, neither of which equals 1 by the work you've already shown
$endgroup$
– cnick
Dec 3 '15 at 1:46
add a comment |
$begingroup$
By factoring $p^2 − 1$, we have $(p + 1)(p - 1)$.
I know that $p = 2$ which gives $3$ is the only solution.
However, how do I prove that $p = 2$ is the only integer which gives a prime?
elementary-number-theory discrete-mathematics proof-verification proof-writing
edited Jul 1 '18 at 20:09
Robert Soupe
11.5k21951
asked Dec 3 '15 at 1:44
Noah DengNoah Deng
8918
$endgroup$
By factoring $p^2 − 1$, we have $(p + 1)(p - 1)$.
I know that $p = 2$ which gives $3$ is the only solution.
However, how do I prove that $p = 2$ is the only integer which gives a prime?
elementary-number-theory discrete-mathematics proof-verification proof-writing
elementary-number-theory discrete-mathematics proof-verification proof-writing
edited Jul 1 '18 at 20:09
Robert Soupe
11.5k21951
asked Dec 3 '15 at 1:44
Noah DengNoah Deng
8918
edited Jul 1 '18 at 20:09
Robert Soupe
11.5k21951
asked Dec 3 '15 at 1:44
Noah DengNoah Deng
8918
edited Jul 1 '18 at 20:09
Robert Soupe
11.5k21951
edited Jul 1 '18 at 20:09
Robert Soupe
11.5k21951
edited Jul 1 '18 at 20:09
Robert Soupe
11.5k21951
11.5k21951
asked Dec 3 '15 at 1:44
Noah DengNoah Deng
8918
asked Dec 3 '15 at 1:44
Noah DengNoah Deng
8918
asked Dec 3 '15 at 1:44
Noah DengNoah Deng
8918
8918
$begingroup$
Because of the $p-1$ factor...
$endgroup$
– abiessu
Dec 3 '15 at 1:45
$begingroup$
for any number>$2$, $p-1$ will give you a factor > $1$
$endgroup$
– Kushal Bhuyan
Dec 3 '15 at 1:46
$begingroup$
any other solution can't be prime because it can be decomposed as a product of two numbers, neither of which equals 1 by the work you've already shown
$endgroup$
– cnick
Dec 3 '15 at 1:46
add a comment |
$begingroup$
Because of the $p-1$ factor...
$endgroup$
– abiessu
Dec 3 '15 at 1:45
$begingroup$
for any number>$2$, $p-1$ will give you a factor > $1$
$endgroup$
– Kushal Bhuyan
Dec 3 '15 at 1:46
$begingroup$
any other solution can't be prime because it can be decomposed as a product of two numbers, neither of which equals 1 by the work you've already shown
$endgroup$
– cnick
Dec 3 '15 at 1:46
$begingroup$
Because of the $p-1$ factor...
$endgroup$
– abiessu
Dec 3 '15 at 1:45
$begingroup$
Because of the $p-1$ factor...
$endgroup$
– abiessu
Dec 3 '15 at 1:45
$begingroup$
for any number>$2$, $p-1$ will give you a factor > $1$
$endgroup$
– Kushal Bhuyan
Dec 3 '15 at 1:46
$begingroup$
for any number>$2$, $p-1$ will give you a factor > $1$
$endgroup$
– Kushal Bhuyan
Dec 3 '15 at 1:46
$begingroup$
any other solution can't be prime because it can be decomposed as a product of two numbers, neither of which equals 1 by the work you've already shown
$endgroup$
– cnick
Dec 3 '15 at 1:46
$begingroup$
any other solution can't be prime because it can be decomposed as a product of two numbers, neither of which equals 1 by the work you've already shown
$endgroup$
– cnick
Dec 3 '15 at 1:46
add a comment |
2 Answers
2
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oldest
votes
$begingroup$
If $p>2$, both $p+1$ and $p-1$ are $>1$, hence $p^2-1$ is composite.
answered Dec 3 '15 at 1:47
BernardBernard
124k742117
$endgroup$
add a comment |
$begingroup$
In $p^2-1=(p-1)(p+1)$ give a prime number only and only if p-1 is 1 so that p+1 will be the only prime factor. So n must be 2 and its the only one which can give us a prime(3).
I better suggest you to think about $$n^2-m=p$$ where 'n' is an natural number and m<$n^2$ such that 'p' will be a prime number.
answered Dec 23 '18 at 13:22
DynamoDynamo
104517
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
If $p>2$, both $p+1$ and $p-1$ are $>1$, hence $p^2-1$ is composite.
answered Dec 3 '15 at 1:47
BernardBernard
124k742117
$endgroup$
add a comment |
$begingroup$
If $p>2$, both $p+1$ and $p-1$ are $>1$, hence $p^2-1$ is composite.
answered Dec 3 '15 at 1:47
BernardBernard
124k742117
$endgroup$
add a comment |
$begingroup$
If $p>2$, both $p+1$ and $p-1$ are $>1$, hence $p^2-1$ is composite.
answered Dec 3 '15 at 1:47
BernardBernard
124k742117
$endgroup$
If $p>2$, both $p+1$ and $p-1$ are $>1$, hence $p^2-1$ is composite.
answered Dec 3 '15 at 1:47
BernardBernard
124k742117
answered Dec 3 '15 at 1:47
BernardBernard
124k742117
answered Dec 3 '15 at 1:47
BernardBernard
124k742117
answered Dec 3 '15 at 1:47
BernardBernard
124k742117
124k742117
add a comment |
add a comment |
$begingroup$
In $p^2-1=(p-1)(p+1)$ give a prime number only and only if p-1 is 1 so that p+1 will be the only prime factor. So n must be 2 and its the only one which can give us a prime(3).
I better suggest you to think about $$n^2-m=p$$ where 'n' is an natural number and m<$n^2$ such that 'p' will be a prime number.
answered Dec 23 '18 at 13:22
DynamoDynamo
104517
$endgroup$
add a comment |
$begingroup$
In $p^2-1=(p-1)(p+1)$ give a prime number only and only if p-1 is 1 so that p+1 will be the only prime factor. So n must be 2 and its the only one which can give us a prime(3).
I better suggest you to think about $$n^2-m=p$$ where 'n' is an natural number and m<$n^2$ such that 'p' will be a prime number.
answered Dec 23 '18 at 13:22
DynamoDynamo
104517
$endgroup$
add a comment |
$begingroup$
In $p^2-1=(p-1)(p+1)$ give a prime number only and only if p-1 is 1 so that p+1 will be the only prime factor. So n must be 2 and its the only one which can give us a prime(3).
I better suggest you to think about $$n^2-m=p$$ where 'n' is an natural number and m<$n^2$ such that 'p' will be a prime number.
answered Dec 23 '18 at 13:22
DynamoDynamo
104517
$endgroup$
In $p^2-1=(p-1)(p+1)$ give a prime number only and only if p-1 is 1 so that p+1 will be the only prime factor. So n must be 2 and its the only one which can give us a prime(3).
I better suggest you to think about $$n^2-m=p$$ where 'n' is an natural number and m<$n^2$ such that 'p' will be a prime number.
answered Dec 23 '18 at 13:22
DynamoDynamo
104517
answered Dec 23 '18 at 13:22
DynamoDynamo
104517
answered Dec 23 '18 at 13:22
DynamoDynamo
104517
answered Dec 23 '18 at 13:22
DynamoDynamo
104517
104517
add a comment |
add a comment |
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$begingroup$
Because of the $p-1$ factor...
$endgroup$
– abiessu
Dec 3 '15 at 1:45
$begingroup$
for any number>$2$, $p-1$ will give you a factor > $1$
$endgroup$
– Kushal Bhuyan
Dec 3 '15 at 1:46
$begingroup$
any other solution can't be prime because it can be decomposed as a product of two numbers, neither of which equals 1 by the work you've already shown
$endgroup$
– cnick
Dec 3 '15 at 1:46