${ninmathbb N|ninmathbb P wedge n ^2+2inmathbb P}$ is finite
In an article on Wikipedia there is a claim of a proof that it don't exist infinitely many $ninmathbb N$ such that both $n$ and $n^2+2$ are primes. I don't understand that and would be pleased if someone could explain.
Okay, thank you, but it is the text in that section that I don't understand. This is supposed to have something to do with integer-valued polynomials and fixed prime divisors.
elementary-number-theory polynomials prime-numbers
asked Jan 5 at 23:14
LehsLehs
6,95231662
add a comment |
In an article on Wikipedia there is a claim of a proof that it don't exist infinitely many $ninmathbb N$ such that both $n$ and $n^2+2$ are primes. I don't understand that and would be pleased if someone could explain.
Okay, thank you, but it is the text in that section that I don't understand. This is supposed to have something to do with integer-valued polynomials and fixed prime divisors.
elementary-number-theory polynomials prime-numbers
asked Jan 5 at 23:14
LehsLehs
6,95231662
2
It's explained on that very page: $3mid n(n^2+2)$.
– Lord Shark the Unknown
Jan 5 at 23:18
4
It's a triviality. If $p>3$ is a prime then $3,|,p^2+2$.
– lulu
Jan 5 at 23:19
add a comment |
In an article on Wikipedia there is a claim of a proof that it don't exist infinitely many $ninmathbb N$ such that both $n$ and $n^2+2$ are primes. I don't understand that and would be pleased if someone could explain.
Okay, thank you, but it is the text in that section that I don't understand. This is supposed to have something to do with integer-valued polynomials and fixed prime divisors.
elementary-number-theory polynomials prime-numbers
asked Jan 5 at 23:14
LehsLehs
6,95231662
In an article on Wikipedia there is a claim of a proof that it don't exist infinitely many $ninmathbb N$ such that both $n$ and $n^2+2$ are primes. I don't understand that and would be pleased if someone could explain.
Okay, thank you, but it is the text in that section that I don't understand. This is supposed to have something to do with integer-valued polynomials and fixed prime divisors.
elementary-number-theory polynomials prime-numbers
elementary-number-theory polynomials prime-numbers
asked Jan 5 at 23:14
LehsLehs
6,95231662
asked Jan 5 at 23:14
LehsLehs
6,95231662
edited 2 days ago
Lehs
asked Jan 5 at 23:14
LehsLehs
6,95231662
asked Jan 5 at 23:14
LehsLehs
6,95231662
asked Jan 5 at 23:14
LehsLehs
6,95231662
6,95231662
2
It's explained on that very page: $3mid n(n^2+2)$.
– Lord Shark the Unknown
Jan 5 at 23:18
4
It's a triviality. If $p>3$ is a prime then $3,|,p^2+2$.
– lulu
Jan 5 at 23:19
add a comment |
2
It's explained on that very page: $3mid n(n^2+2)$.
– Lord Shark the Unknown
Jan 5 at 23:18
4
It's a triviality. If $p>3$ is a prime then $3,|,p^2+2$.
– lulu
Jan 5 at 23:19
2
2
It's explained on that very page: $3mid n(n^2+2)$.
– Lord Shark the Unknown
Jan 5 at 23:18
It's explained on that very page: $3mid n(n^2+2)$.
– Lord Shark the Unknown
Jan 5 at 23:18
4
4
It's a triviality. If $p>3$ is a prime then $3,|,p^2+2$.
– lulu
Jan 5 at 23:19
It's a triviality. If $p>3$ is a prime then $3,|,p^2+2$.
– lulu
Jan 5 at 23:19
add a comment |
6 Answers
6
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If $n$ and $n^2+2$ are both prime, then since $3|n(n^2+2)$, either $3|n$ or $3|n^2+2$. If also $n$ and $n^2+2$ are prime, then either $n = 3$ or $n^2 + 2 = 3$ (so $n=1$, which is not prime). Thus, there is only one (positive) integer prime $n$ such that $n^2+2$ is prime: it is $3$.
answered Jan 5 at 23:21
user3482749user3482749
2,929414
add a comment |
Read the page better: $(n-1)n(n+1)$ is divisible by $3$ as it is a product of three consecutive integers and also $3n$ is divisible by $3$, by definition.
And so $3$ also divides their sum:
$$n(n+1)(n-1) + 3n = n(n^2-1) + 3n=n^3 -n + 3n = n^3 + 2n = n(n^2+2)$$
and so $3$ either divides $n$ or it divides $n^2+2$. So it cannot be that both are primes, except when $n=3$ itself (and we have $3$ and $11$).
answered Jan 5 at 23:23
Henno BrandsmaHenno Brandsma
105k347114
add a comment |
${ninmathbb Nmid ninBbb Pland n^2+2inBbb P}={3}$.
This is because $n(n^2+2)=(n-1)n(n+1)+3n$, and the RHS is divisible by $3$. Then by Euclid's lemma, $3mid nlor3mid(n^2+2)$.
answered Jan 6 at 2:23
Chris CusterChris Custer
11k3824
add a comment |
Consider some prime integer $n geq 4$.
Then we can write $n=3k+r$, where $k$ and $r$ are positive integers ($3$ cannot divide $n$), and $r$ is $1$ or $2$.
Then $n^2+2=(3k+r)^2+2=3(3k^2+2rk)+(r^2+2)$.
You can check that $r^2+2$ is always divisible by $3$; therefore $n^2+2 geq 18$ is divisible by $3$, thus cannot be prime.
edited Jan 6 at 2:28
J. W. Tanner
636
answered Jan 5 at 23:20
MindlackMindlack
2,17717
add a comment |
All primes other than $2,3$ have the form $p=6kpm1$. Accordingly, the squares of those primes have the form $p^2=6j+1$, and $p^2+2$ will have the form $6j+3$ which is divisible by $3$ on its face (see comment by lulu to original question). This leaves $2,3$ as the only possible candidates; $2^2+2=6$ and $3^2+2=11$, making $3$ the sole prime satisfying the condition.
answered Jan 6 at 3:26
Keith BackmanKeith Backman
1,0641712
add a comment |
Now I finally understand what Wikipedia was trying to bring about. The subring of all integer-valued polynomials with rational coefficients is a free Abelian group where all elements $f$ can be uniquely expressed as linear combinations of the (in this ring) irreducible polynomials
$displaystyle left( begin{matrix} X \ k \ end{matrix}right)=
frac{X(X-1)(X-2)cdots(X-k+1)}{k!}$
where $c_0=f(0)$ and $displaystyle c_k=f(k)-sum_{i=0}^{k-1}c_i {kchoose i}$.
Now $|gcd(c_0,dots,c_n)|$ is the greatest number that divides all outputs $f(x)$ for $xinmathbb Z$.
In the example $x(x^2+2)$: $;c_0=0,;c_1=3,;c_2=6,;c_3=6$.
answered 2 days ago
LehsLehs
6,95231662
add a comment |
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6 Answers
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6 Answers
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If $n$ and $n^2+2$ are both prime, then since $3|n(n^2+2)$, either $3|n$ or $3|n^2+2$. If also $n$ and $n^2+2$ are prime, then either $n = 3$ or $n^2 + 2 = 3$ (so $n=1$, which is not prime). Thus, there is only one (positive) integer prime $n$ such that $n^2+2$ is prime: it is $3$.
answered Jan 5 at 23:21
user3482749user3482749
2,929414
add a comment |
If $n$ and $n^2+2$ are both prime, then since $3|n(n^2+2)$, either $3|n$ or $3|n^2+2$. If also $n$ and $n^2+2$ are prime, then either $n = 3$ or $n^2 + 2 = 3$ (so $n=1$, which is not prime). Thus, there is only one (positive) integer prime $n$ such that $n^2+2$ is prime: it is $3$.
answered Jan 5 at 23:21
user3482749user3482749
2,929414
add a comment |
If $n$ and $n^2+2$ are both prime, then since $3|n(n^2+2)$, either $3|n$ or $3|n^2+2$. If also $n$ and $n^2+2$ are prime, then either $n = 3$ or $n^2 + 2 = 3$ (so $n=1$, which is not prime). Thus, there is only one (positive) integer prime $n$ such that $n^2+2$ is prime: it is $3$.
answered Jan 5 at 23:21
user3482749user3482749
2,929414
If $n$ and $n^2+2$ are both prime, then since $3|n(n^2+2)$, either $3|n$ or $3|n^2+2$. If also $n$ and $n^2+2$ are prime, then either $n = 3$ or $n^2 + 2 = 3$ (so $n=1$, which is not prime). Thus, there is only one (positive) integer prime $n$ such that $n^2+2$ is prime: it is $3$.
answered Jan 5 at 23:21
user3482749user3482749
2,929414
answered Jan 5 at 23:21
user3482749user3482749
2,929414
answered Jan 5 at 23:21
user3482749user3482749
2,929414
answered Jan 5 at 23:21
user3482749user3482749
2,929414
2,929414
add a comment |
add a comment |
Read the page better: $(n-1)n(n+1)$ is divisible by $3$ as it is a product of three consecutive integers and also $3n$ is divisible by $3$, by definition.
And so $3$ also divides their sum:
$$n(n+1)(n-1) + 3n = n(n^2-1) + 3n=n^3 -n + 3n = n^3 + 2n = n(n^2+2)$$
and so $3$ either divides $n$ or it divides $n^2+2$. So it cannot be that both are primes, except when $n=3$ itself (and we have $3$ and $11$).
answered Jan 5 at 23:23
Henno BrandsmaHenno Brandsma
105k347114
add a comment |
Read the page better: $(n-1)n(n+1)$ is divisible by $3$ as it is a product of three consecutive integers and also $3n$ is divisible by $3$, by definition.
And so $3$ also divides their sum:
$$n(n+1)(n-1) + 3n = n(n^2-1) + 3n=n^3 -n + 3n = n^3 + 2n = n(n^2+2)$$
and so $3$ either divides $n$ or it divides $n^2+2$. So it cannot be that both are primes, except when $n=3$ itself (and we have $3$ and $11$).
answered Jan 5 at 23:23
Henno BrandsmaHenno Brandsma
105k347114
add a comment |
Read the page better: $(n-1)n(n+1)$ is divisible by $3$ as it is a product of three consecutive integers and also $3n$ is divisible by $3$, by definition.
And so $3$ also divides their sum:
$$n(n+1)(n-1) + 3n = n(n^2-1) + 3n=n^3 -n + 3n = n^3 + 2n = n(n^2+2)$$
and so $3$ either divides $n$ or it divides $n^2+2$. So it cannot be that both are primes, except when $n=3$ itself (and we have $3$ and $11$).
answered Jan 5 at 23:23
Henno BrandsmaHenno Brandsma
105k347114
Read the page better: $(n-1)n(n+1)$ is divisible by $3$ as it is a product of three consecutive integers and also $3n$ is divisible by $3$, by definition.
And so $3$ also divides their sum:
$$n(n+1)(n-1) + 3n = n(n^2-1) + 3n=n^3 -n + 3n = n^3 + 2n = n(n^2+2)$$
and so $3$ either divides $n$ or it divides $n^2+2$. So it cannot be that both are primes, except when $n=3$ itself (and we have $3$ and $11$).
answered Jan 5 at 23:23
Henno BrandsmaHenno Brandsma
105k347114
answered Jan 5 at 23:23
Henno BrandsmaHenno Brandsma
105k347114
answered Jan 5 at 23:23
Henno BrandsmaHenno Brandsma
105k347114
answered Jan 5 at 23:23
Henno BrandsmaHenno Brandsma
105k347114
105k347114
add a comment |
add a comment |
${ninmathbb Nmid ninBbb Pland n^2+2inBbb P}={3}$.
This is because $n(n^2+2)=(n-1)n(n+1)+3n$, and the RHS is divisible by $3$. Then by Euclid's lemma, $3mid nlor3mid(n^2+2)$.
answered Jan 6 at 2:23
Chris CusterChris Custer
11k3824
add a comment |
${ninmathbb Nmid ninBbb Pland n^2+2inBbb P}={3}$.
This is because $n(n^2+2)=(n-1)n(n+1)+3n$, and the RHS is divisible by $3$. Then by Euclid's lemma, $3mid nlor3mid(n^2+2)$.
answered Jan 6 at 2:23
Chris CusterChris Custer
11k3824
add a comment |
${ninmathbb Nmid ninBbb Pland n^2+2inBbb P}={3}$.
This is because $n(n^2+2)=(n-1)n(n+1)+3n$, and the RHS is divisible by $3$. Then by Euclid's lemma, $3mid nlor3mid(n^2+2)$.
answered Jan 6 at 2:23
Chris CusterChris Custer
11k3824
${ninmathbb Nmid ninBbb Pland n^2+2inBbb P}={3}$.
This is because $n(n^2+2)=(n-1)n(n+1)+3n$, and the RHS is divisible by $3$. Then by Euclid's lemma, $3mid nlor3mid(n^2+2)$.
answered Jan 6 at 2:23
Chris CusterChris Custer
11k3824
answered Jan 6 at 2:23
Chris CusterChris Custer
11k3824
answered Jan 6 at 2:23
Chris CusterChris Custer
11k3824
answered Jan 6 at 2:23
Chris CusterChris Custer
11k3824
11k3824
add a comment |
add a comment |
Consider some prime integer $n geq 4$.
Then we can write $n=3k+r$, where $k$ and $r$ are positive integers ($3$ cannot divide $n$), and $r$ is $1$ or $2$.
Then $n^2+2=(3k+r)^2+2=3(3k^2+2rk)+(r^2+2)$.
You can check that $r^2+2$ is always divisible by $3$; therefore $n^2+2 geq 18$ is divisible by $3$, thus cannot be prime.
edited Jan 6 at 2:28
J. W. Tanner
636
answered Jan 5 at 23:20
MindlackMindlack
2,17717
add a comment |
Consider some prime integer $n geq 4$.
Then we can write $n=3k+r$, where $k$ and $r$ are positive integers ($3$ cannot divide $n$), and $r$ is $1$ or $2$.
Then $n^2+2=(3k+r)^2+2=3(3k^2+2rk)+(r^2+2)$.
You can check that $r^2+2$ is always divisible by $3$; therefore $n^2+2 geq 18$ is divisible by $3$, thus cannot be prime.
edited Jan 6 at 2:28
J. W. Tanner
636
answered Jan 5 at 23:20
MindlackMindlack
2,17717
add a comment |
Consider some prime integer $n geq 4$.
Then we can write $n=3k+r$, where $k$ and $r$ are positive integers ($3$ cannot divide $n$), and $r$ is $1$ or $2$.
Then $n^2+2=(3k+r)^2+2=3(3k^2+2rk)+(r^2+2)$.
You can check that $r^2+2$ is always divisible by $3$; therefore $n^2+2 geq 18$ is divisible by $3$, thus cannot be prime.
edited Jan 6 at 2:28
J. W. Tanner
636
answered Jan 5 at 23:20
MindlackMindlack
2,17717
Consider some prime integer $n geq 4$.
Then we can write $n=3k+r$, where $k$ and $r$ are positive integers ($3$ cannot divide $n$), and $r$ is $1$ or $2$.
Then $n^2+2=(3k+r)^2+2=3(3k^2+2rk)+(r^2+2)$.
You can check that $r^2+2$ is always divisible by $3$; therefore $n^2+2 geq 18$ is divisible by $3$, thus cannot be prime.
edited Jan 6 at 2:28
J. W. Tanner
636
answered Jan 5 at 23:20
MindlackMindlack
2,17717
edited Jan 6 at 2:28
J. W. Tanner
636
edited Jan 6 at 2:28
J. W. Tanner
636
edited Jan 6 at 2:28
J. W. Tanner
636
636
answered Jan 5 at 23:20
MindlackMindlack
2,17717
answered Jan 5 at 23:20
MindlackMindlack
2,17717
answered Jan 5 at 23:20
MindlackMindlack
2,17717
2,17717
add a comment |
add a comment |
All primes other than $2,3$ have the form $p=6kpm1$. Accordingly, the squares of those primes have the form $p^2=6j+1$, and $p^2+2$ will have the form $6j+3$ which is divisible by $3$ on its face (see comment by lulu to original question). This leaves $2,3$ as the only possible candidates; $2^2+2=6$ and $3^2+2=11$, making $3$ the sole prime satisfying the condition.
answered Jan 6 at 3:26
Keith BackmanKeith Backman
1,0641712
add a comment |
All primes other than $2,3$ have the form $p=6kpm1$. Accordingly, the squares of those primes have the form $p^2=6j+1$, and $p^2+2$ will have the form $6j+3$ which is divisible by $3$ on its face (see comment by lulu to original question). This leaves $2,3$ as the only possible candidates; $2^2+2=6$ and $3^2+2=11$, making $3$ the sole prime satisfying the condition.
answered Jan 6 at 3:26
Keith BackmanKeith Backman
1,0641712
add a comment |
All primes other than $2,3$ have the form $p=6kpm1$. Accordingly, the squares of those primes have the form $p^2=6j+1$, and $p^2+2$ will have the form $6j+3$ which is divisible by $3$ on its face (see comment by lulu to original question). This leaves $2,3$ as the only possible candidates; $2^2+2=6$ and $3^2+2=11$, making $3$ the sole prime satisfying the condition.
answered Jan 6 at 3:26
Keith BackmanKeith Backman
1,0641712
All primes other than $2,3$ have the form $p=6kpm1$. Accordingly, the squares of those primes have the form $p^2=6j+1$, and $p^2+2$ will have the form $6j+3$ which is divisible by $3$ on its face (see comment by lulu to original question). This leaves $2,3$ as the only possible candidates; $2^2+2=6$ and $3^2+2=11$, making $3$ the sole prime satisfying the condition.
answered Jan 6 at 3:26
Keith BackmanKeith Backman
1,0641712
answered Jan 6 at 3:26
Keith BackmanKeith Backman
1,0641712
answered Jan 6 at 3:26
Keith BackmanKeith Backman
1,0641712
answered Jan 6 at 3:26
Keith BackmanKeith Backman
1,0641712
1,0641712
add a comment |
add a comment |
Now I finally understand what Wikipedia was trying to bring about. The subring of all integer-valued polynomials with rational coefficients is a free Abelian group where all elements $f$ can be uniquely expressed as linear combinations of the (in this ring) irreducible polynomials
$displaystyle left( begin{matrix} X \ k \ end{matrix}right)=
frac{X(X-1)(X-2)cdots(X-k+1)}{k!}$
where $c_0=f(0)$ and $displaystyle c_k=f(k)-sum_{i=0}^{k-1}c_i {kchoose i}$.
Now $|gcd(c_0,dots,c_n)|$ is the greatest number that divides all outputs $f(x)$ for $xinmathbb Z$.
In the example $x(x^2+2)$: $;c_0=0,;c_1=3,;c_2=6,;c_3=6$.
answered 2 days ago
LehsLehs
6,95231662
add a comment |
Now I finally understand what Wikipedia was trying to bring about. The subring of all integer-valued polynomials with rational coefficients is a free Abelian group where all elements $f$ can be uniquely expressed as linear combinations of the (in this ring) irreducible polynomials
$displaystyle left( begin{matrix} X \ k \ end{matrix}right)=
frac{X(X-1)(X-2)cdots(X-k+1)}{k!}$
where $c_0=f(0)$ and $displaystyle c_k=f(k)-sum_{i=0}^{k-1}c_i {kchoose i}$.
Now $|gcd(c_0,dots,c_n)|$ is the greatest number that divides all outputs $f(x)$ for $xinmathbb Z$.
In the example $x(x^2+2)$: $;c_0=0,;c_1=3,;c_2=6,;c_3=6$.
answered 2 days ago
LehsLehs
6,95231662
add a comment |
Now I finally understand what Wikipedia was trying to bring about. The subring of all integer-valued polynomials with rational coefficients is a free Abelian group where all elements $f$ can be uniquely expressed as linear combinations of the (in this ring) irreducible polynomials
$displaystyle left( begin{matrix} X \ k \ end{matrix}right)=
frac{X(X-1)(X-2)cdots(X-k+1)}{k!}$
where $c_0=f(0)$ and $displaystyle c_k=f(k)-sum_{i=0}^{k-1}c_i {kchoose i}$.
Now $|gcd(c_0,dots,c_n)|$ is the greatest number that divides all outputs $f(x)$ for $xinmathbb Z$.
In the example $x(x^2+2)$: $;c_0=0,;c_1=3,;c_2=6,;c_3=6$.
answered 2 days ago
LehsLehs
6,95231662
Now I finally understand what Wikipedia was trying to bring about. The subring of all integer-valued polynomials with rational coefficients is a free Abelian group where all elements $f$ can be uniquely expressed as linear combinations of the (in this ring) irreducible polynomials
$displaystyle left( begin{matrix} X \ k \ end{matrix}right)=
frac{X(X-1)(X-2)cdots(X-k+1)}{k!}$
where $c_0=f(0)$ and $displaystyle c_k=f(k)-sum_{i=0}^{k-1}c_i {kchoose i}$.
Now $|gcd(c_0,dots,c_n)|$ is the greatest number that divides all outputs $f(x)$ for $xinmathbb Z$.
In the example $x(x^2+2)$: $;c_0=0,;c_1=3,;c_2=6,;c_3=6$.
answered 2 days ago
LehsLehs
6,95231662
edited 2 days ago
answered 2 days ago
LehsLehs
6,95231662
answered 2 days ago
LehsLehs
6,95231662
answered 2 days ago
LehsLehs
6,95231662
6,95231662
add a comment |
add a comment |
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2
It's explained on that very page: $3mid n(n^2+2)$.
– Lord Shark the Unknown
Jan 5 at 23:18
4
It's a triviality. If $p>3$ is a prime then $3,|,p^2+2$.
– lulu
Jan 5 at 23:19