Prove that there exists a unique prime number of the form $n^2 − 1$, where *n* is an integer that is greater than or equal to 2.Prove that n!+1 contains a prime factor greater than n and use this to prove that there are infinte many primesshow that $q$ and $r$ are unique when $r$ is less than or equal to zero.Proof of prime numbers in the form..Prove that there exists only one prime number of the form $p^2−1$ where $p≥2$ is an integerProof that there exists a larger prime than prime number P, which is the largest of a finite set of primes?Prove by contradiction: Let n > 1. Let a be the smallest factor of n that is greater than 1. Prove a is prime.Strong Induction - Any integer greater than 1 is divisible by a prime numberProve that the square root of a positive integer is either an integer or irrationalProve or Disprove that there exists an integer $n$ such that $4n^2 -12n +8$ is primeShow that given any rational number $x$, there exists an integer $y$ such that $x^2y$ is an integer
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Prove that there exists a unique prime number of the form $n^2 − 1$, where *n* is an integer that is greater than or equal to 2.
Prove that n!+1 contains a prime factor greater than n and use this to prove that there are infinte many primesshow that $q$ and $r$ are unique when $r$ is less than or equal to zero.Proof of prime numbers in the form..Prove that there exists only one prime number of the form $p^2−1$ where $p≥2$ is an integerProof that there exists a larger prime than prime number P, which is the largest of a finite set of primes?Prove by contradiction: Let n > 1. Let a be the smallest factor of n that is greater than 1. Prove a is prime.Strong Induction - Any integer greater than 1 is divisible by a prime numberProve that the square root of a positive integer is either an integer or irrationalProve or Disprove that there exists an integer $n$ such that $4n^2 -12n +8$ is primeShow that given any rational number $x$, there exists an integer $y$ such that $x^2y$ is an integer
$begingroup$
Prove that there exists a unique prime number of the form $n^2-1$,
where n is an integer that is greater than or equal to 2.
I know I can factor $n^2-1 = (n-1)(n+1)$. What are my next steps?
discrete-mathematics proof-explanation
asked Mar 12 at 20:32
Usama GhawjiUsama Ghawji
666
$endgroup$
add a comment |
$begingroup$
Prove that there exists a unique prime number of the form $n^2-1$,
where n is an integer that is greater than or equal to 2.
I know I can factor $n^2-1 = (n-1)(n+1)$. What are my next steps?
discrete-mathematics proof-explanation
asked Mar 12 at 20:32
Usama GhawjiUsama Ghawji
666
$endgroup$
1
$begingroup$
n-1=1 and n+1=p because p is prime and n-1<n+1
$endgroup$
– Anthony
Mar 12 at 20:33
add a comment |
$begingroup$
Prove that there exists a unique prime number of the form $n^2-1$,
where n is an integer that is greater than or equal to 2.
I know I can factor $n^2-1 = (n-1)(n+1)$. What are my next steps?
discrete-mathematics proof-explanation
asked Mar 12 at 20:32
Usama GhawjiUsama Ghawji
666
$endgroup$
Prove that there exists a unique prime number of the form $n^2-1$,
where n is an integer that is greater than or equal to 2.
I know I can factor $n^2-1 = (n-1)(n+1)$. What are my next steps?
discrete-mathematics proof-explanation
discrete-mathematics proof-explanation
asked Mar 12 at 20:32
Usama GhawjiUsama Ghawji
666
asked Mar 12 at 20:32
Usama GhawjiUsama Ghawji
666
asked Mar 12 at 20:32
Usama GhawjiUsama Ghawji
666
asked Mar 12 at 20:32
Usama GhawjiUsama Ghawji
666
asked Mar 12 at 20:32
Usama GhawjiUsama Ghawji
666
666
1
$begingroup$
n-1=1 and n+1=p because p is prime and n-1<n+1
$endgroup$
– Anthony
Mar 12 at 20:33
add a comment |
1
$begingroup$
n-1=1 and n+1=p because p is prime and n-1<n+1
$endgroup$
– Anthony
Mar 12 at 20:33
1
1
$begingroup$
n-1=1 and n+1=p because p is prime and n-1<n+1
$endgroup$
– Anthony
Mar 12 at 20:33
$begingroup$
n-1=1 and n+1=p because p is prime and n-1<n+1
$endgroup$
– Anthony
Mar 12 at 20:33
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Existence : $2^2 - 1 = 3$
Uniqueness : Suppose we have a number $n > 2$ with prime number of the form $n^2 - 1$. Then since $n^2 - 1 = (n+1)cdot (n-1)$, we have $(n+1) > 1$ and $(n-1) > 1$, which means that both of these are factors that are not $1$ and thus $n^2-1$ is not prime.
answered Mar 12 at 20:36
twnlytwnly
1,1891213
$endgroup$
$begingroup$
Thank you I appreciate your explanation.
$endgroup$
– Usama Ghawji
Mar 12 at 20:53
add a comment |
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$begingroup$
Existence : $2^2 - 1 = 3$
Uniqueness : Suppose we have a number $n > 2$ with prime number of the form $n^2 - 1$. Then since $n^2 - 1 = (n+1)cdot (n-1)$, we have $(n+1) > 1$ and $(n-1) > 1$, which means that both of these are factors that are not $1$ and thus $n^2-1$ is not prime.
answered Mar 12 at 20:36
twnlytwnly
1,1891213
$endgroup$
$begingroup$
Thank you I appreciate your explanation.
$endgroup$
– Usama Ghawji
Mar 12 at 20:53
add a comment |
$begingroup$
Existence : $2^2 - 1 = 3$
Uniqueness : Suppose we have a number $n > 2$ with prime number of the form $n^2 - 1$. Then since $n^2 - 1 = (n+1)cdot (n-1)$, we have $(n+1) > 1$ and $(n-1) > 1$, which means that both of these are factors that are not $1$ and thus $n^2-1$ is not prime.
answered Mar 12 at 20:36
twnlytwnly
1,1891213
$endgroup$
$begingroup$
Thank you I appreciate your explanation.
$endgroup$
– Usama Ghawji
Mar 12 at 20:53
add a comment |
$begingroup$
Existence : $2^2 - 1 = 3$
Uniqueness : Suppose we have a number $n > 2$ with prime number of the form $n^2 - 1$. Then since $n^2 - 1 = (n+1)cdot (n-1)$, we have $(n+1) > 1$ and $(n-1) > 1$, which means that both of these are factors that are not $1$ and thus $n^2-1$ is not prime.
answered Mar 12 at 20:36
twnlytwnly
1,1891213
$endgroup$
Existence : $2^2 - 1 = 3$
Uniqueness : Suppose we have a number $n > 2$ with prime number of the form $n^2 - 1$. Then since $n^2 - 1 = (n+1)cdot (n-1)$, we have $(n+1) > 1$ and $(n-1) > 1$, which means that both of these are factors that are not $1$ and thus $n^2-1$ is not prime.
answered Mar 12 at 20:36
twnlytwnly
1,1891213
answered Mar 12 at 20:36
twnlytwnly
1,1891213
answered Mar 12 at 20:36
twnlytwnly
1,1891213
answered Mar 12 at 20:36
twnlytwnly
1,1891213
1,1891213
$begingroup$
Thank you I appreciate your explanation.
$endgroup$
– Usama Ghawji
Mar 12 at 20:53
add a comment |
$begingroup$
Thank you I appreciate your explanation.
$endgroup$
– Usama Ghawji
Mar 12 at 20:53
$begingroup$
Thank you I appreciate your explanation.
$endgroup$
– Usama Ghawji
Mar 12 at 20:53
$begingroup$
Thank you I appreciate your explanation.
$endgroup$
– Usama Ghawji
Mar 12 at 20:53
add a comment |
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$begingroup$
n-1=1 and n+1=p because p is prime and n-1<n+1
$endgroup$
– Anthony
Mar 12 at 20:33