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Proving that there are infinitely-many prime numbers that are not Fibonacci numbers
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Fibonacci numbers ${F_n}_ninmathbb{N}$ are defined by the sequence $F_{n+2}=F_{n+1}+F_{n}$, $F_1=F_2=1$. Now prove that there are infinitely many prime numbers which are not Fibonacci numbers.
I tried very much and tried to get a proof by contradiction, but failed. I assumed the negation, but it takes me nowhere.
I assumed that there is a prime number which is a Fibonacci number and the next prime is also a Fibonacci number. Hence, as there is at least one prime between $n$ and $2n$, then $2n$ will have a Fibonacci number form, which can be hence expressed as a prime. But I am not sure what to do next. Is my assumption correct after all?
number-theory fibonacci-numbers
edited Jan 20 at 15:20
Thomas Shelby
3,7492525
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add a comment |
$begingroup$
Fibonacci numbers ${F_n}_ninmathbb{N}$ are defined by the sequence $F_{n+2}=F_{n+1}+F_{n}$, $F_1=F_2=1$. Now prove that there are infinitely many prime numbers which are not Fibonacci numbers.
I tried very much and tried to get a proof by contradiction, but failed. I assumed the negation, but it takes me nowhere.
I assumed that there is a prime number which is a Fibonacci number and the next prime is also a Fibonacci number. Hence, as there is at least one prime between $n$ and $2n$, then $2n$ will have a Fibonacci number form, which can be hence expressed as a prime. But I am not sure what to do next. Is my assumption correct after all?
number-theory fibonacci-numbers
edited Jan 20 at 15:20
Thomas Shelby
3,7492525
$endgroup$
$begingroup$
Why the negative reactions ? The question is clear and an effort is shown.
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– Peter
Jan 20 at 14:48
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yes that is what i am thinking
$endgroup$
– user636268
Jan 20 at 14:52
add a comment |
$begingroup$
Fibonacci numbers ${F_n}_ninmathbb{N}$ are defined by the sequence $F_{n+2}=F_{n+1}+F_{n}$, $F_1=F_2=1$. Now prove that there are infinitely many prime numbers which are not Fibonacci numbers.
I tried very much and tried to get a proof by contradiction, but failed. I assumed the negation, but it takes me nowhere.
I assumed that there is a prime number which is a Fibonacci number and the next prime is also a Fibonacci number. Hence, as there is at least one prime between $n$ and $2n$, then $2n$ will have a Fibonacci number form, which can be hence expressed as a prime. But I am not sure what to do next. Is my assumption correct after all?
number-theory fibonacci-numbers
edited Jan 20 at 15:20
Thomas Shelby
3,7492525
$endgroup$
Fibonacci numbers ${F_n}_ninmathbb{N}$ are defined by the sequence $F_{n+2}=F_{n+1}+F_{n}$, $F_1=F_2=1$. Now prove that there are infinitely many prime numbers which are not Fibonacci numbers.
I tried very much and tried to get a proof by contradiction, but failed. I assumed the negation, but it takes me nowhere.
I assumed that there is a prime number which is a Fibonacci number and the next prime is also a Fibonacci number. Hence, as there is at least one prime between $n$ and $2n$, then $2n$ will have a Fibonacci number form, which can be hence expressed as a prime. But I am not sure what to do next. Is my assumption correct after all?
number-theory fibonacci-numbers
number-theory fibonacci-numbers
edited Jan 20 at 15:20
Thomas Shelby
3,7492525
edited Jan 20 at 15:20
Thomas Shelby
3,7492525
edited Jan 20 at 15:20
Thomas Shelby
3,7492525
edited Jan 20 at 15:20
Thomas Shelby
3,7492525
edited Jan 20 at 15:20
Thomas Shelby
3,7492525
3,7492525
asked Jan 20 at 14:35
user636268
$begingroup$
Why the negative reactions ? The question is clear and an effort is shown.
$endgroup$
– Peter
Jan 20 at 14:48
$begingroup$
yes that is what i am thinking
$endgroup$
– user636268
Jan 20 at 14:52
add a comment |
$begingroup$
Why the negative reactions ? The question is clear and an effort is shown.
$endgroup$
– Peter
Jan 20 at 14:48
$begingroup$
yes that is what i am thinking
$endgroup$
– user636268
Jan 20 at 14:52
$begingroup$
Why the negative reactions ? The question is clear and an effort is shown.
$endgroup$
– Peter
Jan 20 at 14:48
$begingroup$
Why the negative reactions ? The question is clear and an effort is shown.
$endgroup$
– Peter
Jan 20 at 14:48
$begingroup$
yes that is what i am thinking
$endgroup$
– user636268
Jan 20 at 14:52
$begingroup$
yes that is what i am thinking
$endgroup$
– user636268
Jan 20 at 14:52
add a comment |
1 Answer
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Analyzing mod $11$, you can show that no fibonacci-number has the form $11k+4$, but infinite many primes have that form because of Dirichlet's theorem.
answered Jan 20 at 14:44
PeterPeter
48.2k1139133
$endgroup$
$begingroup$
One can easily prove (by induction or otherwise) the identities $$F_{2n} = (F_{n-1} + F_{n+1}) F_n,$$ $$F_{2n-1} = F^2_n + F^2_{n-1}.$$ From the first identity, $F_{2n}$ is never prime for $n > 2$. From the second identity, if $F_{2n-1} = p$ is prime, then it is a sum of squares, which forces $p = 2$ or $p equiv 1 bmod 4$. So the only prime $F_n$ are either $F_3 = 2$, $F_4 = 3$, or are $1 bmod 4$. But there are infinitely many primes of the form $3 bmod 4$ (take $4$ times the product of all such primes and then subtract one; this number has a new prime factor of this form).
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– Lorem Ipsum
Jan 21 at 0:14
$begingroup$
This is similar to the answer given, but requires a (much) easier case of Dirichlet's Theorem.
$endgroup$
– Lorem Ipsum
Jan 21 at 0:14
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Analyzing mod $11$, you can show that no fibonacci-number has the form $11k+4$, but infinite many primes have that form because of Dirichlet's theorem.
answered Jan 20 at 14:44
PeterPeter
48.2k1139133
$endgroup$
$begingroup$
One can easily prove (by induction or otherwise) the identities $$F_{2n} = (F_{n-1} + F_{n+1}) F_n,$$ $$F_{2n-1} = F^2_n + F^2_{n-1}.$$ From the first identity, $F_{2n}$ is never prime for $n > 2$. From the second identity, if $F_{2n-1} = p$ is prime, then it is a sum of squares, which forces $p = 2$ or $p equiv 1 bmod 4$. So the only prime $F_n$ are either $F_3 = 2$, $F_4 = 3$, or are $1 bmod 4$. But there are infinitely many primes of the form $3 bmod 4$ (take $4$ times the product of all such primes and then subtract one; this number has a new prime factor of this form).
$endgroup$
– Lorem Ipsum
Jan 21 at 0:14
$begingroup$
This is similar to the answer given, but requires a (much) easier case of Dirichlet's Theorem.
$endgroup$
– Lorem Ipsum
Jan 21 at 0:14
add a comment |
$begingroup$
Analyzing mod $11$, you can show that no fibonacci-number has the form $11k+4$, but infinite many primes have that form because of Dirichlet's theorem.
answered Jan 20 at 14:44
PeterPeter
48.2k1139133
$endgroup$
$begingroup$
One can easily prove (by induction or otherwise) the identities $$F_{2n} = (F_{n-1} + F_{n+1}) F_n,$$ $$F_{2n-1} = F^2_n + F^2_{n-1}.$$ From the first identity, $F_{2n}$ is never prime for $n > 2$. From the second identity, if $F_{2n-1} = p$ is prime, then it is a sum of squares, which forces $p = 2$ or $p equiv 1 bmod 4$. So the only prime $F_n$ are either $F_3 = 2$, $F_4 = 3$, or are $1 bmod 4$. But there are infinitely many primes of the form $3 bmod 4$ (take $4$ times the product of all such primes and then subtract one; this number has a new prime factor of this form).
$endgroup$
– Lorem Ipsum
Jan 21 at 0:14
$begingroup$
This is similar to the answer given, but requires a (much) easier case of Dirichlet's Theorem.
$endgroup$
– Lorem Ipsum
Jan 21 at 0:14
add a comment |
$begingroup$
Analyzing mod $11$, you can show that no fibonacci-number has the form $11k+4$, but infinite many primes have that form because of Dirichlet's theorem.
answered Jan 20 at 14:44
PeterPeter
48.2k1139133
$endgroup$
Analyzing mod $11$, you can show that no fibonacci-number has the form $11k+4$, but infinite many primes have that form because of Dirichlet's theorem.
answered Jan 20 at 14:44
PeterPeter
48.2k1139133
answered Jan 20 at 14:44
PeterPeter
48.2k1139133
answered Jan 20 at 14:44
PeterPeter
48.2k1139133
answered Jan 20 at 14:44
PeterPeter
48.2k1139133
48.2k1139133
$begingroup$
One can easily prove (by induction or otherwise) the identities $$F_{2n} = (F_{n-1} + F_{n+1}) F_n,$$ $$F_{2n-1} = F^2_n + F^2_{n-1}.$$ From the first identity, $F_{2n}$ is never prime for $n > 2$. From the second identity, if $F_{2n-1} = p$ is prime, then it is a sum of squares, which forces $p = 2$ or $p equiv 1 bmod 4$. So the only prime $F_n$ are either $F_3 = 2$, $F_4 = 3$, or are $1 bmod 4$. But there are infinitely many primes of the form $3 bmod 4$ (take $4$ times the product of all such primes and then subtract one; this number has a new prime factor of this form).
$endgroup$
– Lorem Ipsum
Jan 21 at 0:14
$begingroup$
This is similar to the answer given, but requires a (much) easier case of Dirichlet's Theorem.
$endgroup$
– Lorem Ipsum
Jan 21 at 0:14
add a comment |
$begingroup$
One can easily prove (by induction or otherwise) the identities $$F_{2n} = (F_{n-1} + F_{n+1}) F_n,$$ $$F_{2n-1} = F^2_n + F^2_{n-1}.$$ From the first identity, $F_{2n}$ is never prime for $n > 2$. From the second identity, if $F_{2n-1} = p$ is prime, then it is a sum of squares, which forces $p = 2$ or $p equiv 1 bmod 4$. So the only prime $F_n$ are either $F_3 = 2$, $F_4 = 3$, or are $1 bmod 4$. But there are infinitely many primes of the form $3 bmod 4$ (take $4$ times the product of all such primes and then subtract one; this number has a new prime factor of this form).
$endgroup$
– Lorem Ipsum
Jan 21 at 0:14
$begingroup$
This is similar to the answer given, but requires a (much) easier case of Dirichlet's Theorem.
$endgroup$
– Lorem Ipsum
Jan 21 at 0:14
$begingroup$
One can easily prove (by induction or otherwise) the identities $$F_{2n} = (F_{n-1} + F_{n+1}) F_n,$$ $$F_{2n-1} = F^2_n + F^2_{n-1}.$$ From the first identity, $F_{2n}$ is never prime for $n > 2$. From the second identity, if $F_{2n-1} = p$ is prime, then it is a sum of squares, which forces $p = 2$ or $p equiv 1 bmod 4$. So the only prime $F_n$ are either $F_3 = 2$, $F_4 = 3$, or are $1 bmod 4$. But there are infinitely many primes of the form $3 bmod 4$ (take $4$ times the product of all such primes and then subtract one; this number has a new prime factor of this form).
$endgroup$
– Lorem Ipsum
Jan 21 at 0:14
$begingroup$
One can easily prove (by induction or otherwise) the identities $$F_{2n} = (F_{n-1} + F_{n+1}) F_n,$$ $$F_{2n-1} = F^2_n + F^2_{n-1}.$$ From the first identity, $F_{2n}$ is never prime for $n > 2$. From the second identity, if $F_{2n-1} = p$ is prime, then it is a sum of squares, which forces $p = 2$ or $p equiv 1 bmod 4$. So the only prime $F_n$ are either $F_3 = 2$, $F_4 = 3$, or are $1 bmod 4$. But there are infinitely many primes of the form $3 bmod 4$ (take $4$ times the product of all such primes and then subtract one; this number has a new prime factor of this form).
$endgroup$
– Lorem Ipsum
Jan 21 at 0:14
$begingroup$
This is similar to the answer given, but requires a (much) easier case of Dirichlet's Theorem.
$endgroup$
– Lorem Ipsum
Jan 21 at 0:14
$begingroup$
This is similar to the answer given, but requires a (much) easier case of Dirichlet's Theorem.
$endgroup$
– Lorem Ipsum
Jan 21 at 0:14
add a comment |
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$begingroup$
Why the negative reactions ? The question is clear and an effort is shown.
$endgroup$
– Peter
Jan 20 at 14:48
$begingroup$
yes that is what i am thinking
$endgroup$
– user636268
Jan 20 at 14:52