A mysterious connection between primes and squares
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17
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Motivated by two previous questions of mine (cf. Primes arising from permutations and Primes arising from permutations (II)), here I ask a curious question which connects primes with squares.
QUESTION: Is my following conjecture true?
Conjecture. Let $n$ be any positive integer, and let $S(n)$ denote the number of permutations $tau$ of ${1,ldots,n}$ with $k^4+tau(k)^4$ prime for all $k=1,ldots,n$. Then $S(n)$ is always a positive square.
Via a computer, I find that
$$(S(1),ldots,S(11))=(1,1,1,4,4,4,4,64,16,144,144).$$
For example, $(1,3,2)$ is a permutation of ${1,2,3}$ with
$$1^4+1^4=2, 2^4+3^4=97, text{and} 3^4+2^4=97$$
all prime.
nt.number-theory co.combinatorics prime-numbers permutations
asked Nov 15 at 4:36
Zhi-Wei Sun
2,229223
|
show 1 more comment
up vote
17
down vote
favorite
Motivated by two previous questions of mine (cf. Primes arising from permutations and Primes arising from permutations (II)), here I ask a curious question which connects primes with squares.
QUESTION: Is my following conjecture true?
Conjecture. Let $n$ be any positive integer, and let $S(n)$ denote the number of permutations $tau$ of ${1,ldots,n}$ with $k^4+tau(k)^4$ prime for all $k=1,ldots,n$. Then $S(n)$ is always a positive square.
Via a computer, I find that
$$(S(1),ldots,S(11))=(1,1,1,4,4,4,4,64,16,144,144).$$
For example, $(1,3,2)$ is a permutation of ${1,2,3}$ with
$$1^4+1^4=2, 2^4+3^4=97, text{and} 3^4+2^4=97$$
all prime.
nt.number-theory co.combinatorics prime-numbers permutations
asked Nov 15 at 4:36
Zhi-Wei Sun
2,229223
9
$(S(12),dots,S(20))=(0,12^2,12^2,17^2,66^2,54^2,150^2,282^2,1374^2)$. I computed the permanent of the matrix $M$ where $M_{ij}$ is $1$ if $i^4+j^4$ is prime and $0$ otherwise.
– MTyson
Nov 15 at 6:02
7
Note that $M$ is symmetric and that all $1$'s are in entries where $i+j$ is odd, except for the $1$ in the upper left. Therefore by sorting rows and columns into the order $(1,3,5,dots,2,4,6,dots)$ it can be transformed into a block matrix $begin{pmatrix}W&X\X^top& 0end{pmatrix}$, where $W$ is $0$ except for a $1$ in the upper left entry. If it weren't for this $1$, we would have $mathrm{perm}(M)=mathrm{perm}(X)^2$.
– MTyson
Nov 15 at 6:29
2
@MTyson it doesn't matter if $W$ has $1$ in the upper left entry or has $0$. $mathrm{perm}(M)=mathrm{perm}(X)^2$ still holds.
– Nemo
Nov 15 at 7:38
1
@Nemo When $n$ is odd, $X$ is not even a square matrix...
– Zach Teitler
Nov 15 at 11:17
1
I think this conjecture was hasty and both @MTyson comments and Ilya Bogdanov's answer show the result has very little to do with primes.
– Yaakov Baruch
Nov 16 at 9:45
|
show 1 more comment
up vote
17
down vote
favorite
up vote
17
down vote
favorite
Motivated by two previous questions of mine (cf. Primes arising from permutations and Primes arising from permutations (II)), here I ask a curious question which connects primes with squares.
QUESTION: Is my following conjecture true?
Conjecture. Let $n$ be any positive integer, and let $S(n)$ denote the number of permutations $tau$ of ${1,ldots,n}$ with $k^4+tau(k)^4$ prime for all $k=1,ldots,n$. Then $S(n)$ is always a positive square.
Via a computer, I find that
$$(S(1),ldots,S(11))=(1,1,1,4,4,4,4,64,16,144,144).$$
For example, $(1,3,2)$ is a permutation of ${1,2,3}$ with
$$1^4+1^4=2, 2^4+3^4=97, text{and} 3^4+2^4=97$$
all prime.
nt.number-theory co.combinatorics prime-numbers permutations
asked Nov 15 at 4:36
Zhi-Wei Sun
2,229223
Motivated by two previous questions of mine (cf. Primes arising from permutations and Primes arising from permutations (II)), here I ask a curious question which connects primes with squares.
QUESTION: Is my following conjecture true?
Conjecture. Let $n$ be any positive integer, and let $S(n)$ denote the number of permutations $tau$ of ${1,ldots,n}$ with $k^4+tau(k)^4$ prime for all $k=1,ldots,n$. Then $S(n)$ is always a positive square.
Via a computer, I find that
$$(S(1),ldots,S(11))=(1,1,1,4,4,4,4,64,16,144,144).$$
For example, $(1,3,2)$ is a permutation of ${1,2,3}$ with
$$1^4+1^4=2, 2^4+3^4=97, text{and} 3^4+2^4=97$$
all prime.
nt.number-theory co.combinatorics prime-numbers permutations
nt.number-theory co.combinatorics prime-numbers permutations
asked Nov 15 at 4:36
Zhi-Wei Sun
2,229223
asked Nov 15 at 4:36
Zhi-Wei Sun
2,229223
asked Nov 15 at 4:36
Zhi-Wei Sun
2,229223
asked Nov 15 at 4:36
Zhi-Wei Sun
2,229223
asked Nov 15 at 4:36
Zhi-Wei Sun
2,229223
2,229223
9
$(S(12),dots,S(20))=(0,12^2,12^2,17^2,66^2,54^2,150^2,282^2,1374^2)$. I computed the permanent of the matrix $M$ where $M_{ij}$ is $1$ if $i^4+j^4$ is prime and $0$ otherwise.
– MTyson
Nov 15 at 6:02
7
Note that $M$ is symmetric and that all $1$'s are in entries where $i+j$ is odd, except for the $1$ in the upper left. Therefore by sorting rows and columns into the order $(1,3,5,dots,2,4,6,dots)$ it can be transformed into a block matrix $begin{pmatrix}W&X\X^top& 0end{pmatrix}$, where $W$ is $0$ except for a $1$ in the upper left entry. If it weren't for this $1$, we would have $mathrm{perm}(M)=mathrm{perm}(X)^2$.
– MTyson
Nov 15 at 6:29
2
@MTyson it doesn't matter if $W$ has $1$ in the upper left entry or has $0$. $mathrm{perm}(M)=mathrm{perm}(X)^2$ still holds.
– Nemo
Nov 15 at 7:38
1
@Nemo When $n$ is odd, $X$ is not even a square matrix...
– Zach Teitler
Nov 15 at 11:17
1
I think this conjecture was hasty and both @MTyson comments and Ilya Bogdanov's answer show the result has very little to do with primes.
– Yaakov Baruch
Nov 16 at 9:45
|
show 1 more comment
9
$(S(12),dots,S(20))=(0,12^2,12^2,17^2,66^2,54^2,150^2,282^2,1374^2)$. I computed the permanent of the matrix $M$ where $M_{ij}$ is $1$ if $i^4+j^4$ is prime and $0$ otherwise.
– MTyson
Nov 15 at 6:02
7
Note that $M$ is symmetric and that all $1$'s are in entries where $i+j$ is odd, except for the $1$ in the upper left. Therefore by sorting rows and columns into the order $(1,3,5,dots,2,4,6,dots)$ it can be transformed into a block matrix $begin{pmatrix}W&X\X^top& 0end{pmatrix}$, where $W$ is $0$ except for a $1$ in the upper left entry. If it weren't for this $1$, we would have $mathrm{perm}(M)=mathrm{perm}(X)^2$.
– MTyson
Nov 15 at 6:29
2
@MTyson it doesn't matter if $W$ has $1$ in the upper left entry or has $0$. $mathrm{perm}(M)=mathrm{perm}(X)^2$ still holds.
– Nemo
Nov 15 at 7:38
1
@Nemo When $n$ is odd, $X$ is not even a square matrix...
– Zach Teitler
Nov 15 at 11:17
1
I think this conjecture was hasty and both @MTyson comments and Ilya Bogdanov's answer show the result has very little to do with primes.
– Yaakov Baruch
Nov 16 at 9:45
9
9
$(S(12),dots,S(20))=(0,12^2,12^2,17^2,66^2,54^2,150^2,282^2,1374^2)$. I computed the permanent of the matrix $M$ where $M_{ij}$ is $1$ if $i^4+j^4$ is prime and $0$ otherwise.
– MTyson
Nov 15 at 6:02
$(S(12),dots,S(20))=(0,12^2,12^2,17^2,66^2,54^2,150^2,282^2,1374^2)$. I computed the permanent of the matrix $M$ where $M_{ij}$ is $1$ if $i^4+j^4$ is prime and $0$ otherwise.
– MTyson
Nov 15 at 6:02
7
7
Note that $M$ is symmetric and that all $1$'s are in entries where $i+j$ is odd, except for the $1$ in the upper left. Therefore by sorting rows and columns into the order $(1,3,5,dots,2,4,6,dots)$ it can be transformed into a block matrix $begin{pmatrix}W&X\X^top& 0end{pmatrix}$, where $W$ is $0$ except for a $1$ in the upper left entry. If it weren't for this $1$, we would have $mathrm{perm}(M)=mathrm{perm}(X)^2$.
– MTyson
Nov 15 at 6:29
Note that $M$ is symmetric and that all $1$'s are in entries where $i+j$ is odd, except for the $1$ in the upper left. Therefore by sorting rows and columns into the order $(1,3,5,dots,2,4,6,dots)$ it can be transformed into a block matrix $begin{pmatrix}W&X\X^top& 0end{pmatrix}$, where $W$ is $0$ except for a $1$ in the upper left entry. If it weren't for this $1$, we would have $mathrm{perm}(M)=mathrm{perm}(X)^2$.
– MTyson
Nov 15 at 6:29
2
2
@MTyson it doesn't matter if $W$ has $1$ in the upper left entry or has $0$. $mathrm{perm}(M)=mathrm{perm}(X)^2$ still holds.
– Nemo
Nov 15 at 7:38
@MTyson it doesn't matter if $W$ has $1$ in the upper left entry or has $0$. $mathrm{perm}(M)=mathrm{perm}(X)^2$ still holds.
– Nemo
Nov 15 at 7:38
1
1
@Nemo When $n$ is odd, $X$ is not even a square matrix...
– Zach Teitler
Nov 15 at 11:17
@Nemo When $n$ is odd, $X$ is not even a square matrix...
– Zach Teitler
Nov 15 at 11:17
1
1
I think this conjecture was hasty and both @MTyson comments and Ilya Bogdanov's answer show the result has very little to do with primes.
– Yaakov Baruch
Nov 16 at 9:45
I think this conjecture was hasty and both @MTyson comments and Ilya Bogdanov's answer show the result has very little to do with primes.
– Yaakov Baruch
Nov 16 at 9:45
|
show 1 more comment
1 Answer
1
active
oldest
votes
up vote
21
down vote
The fact on squareness is easy.
If $n$ is odd, among the sums of the form $i^4+tau_i^4$ there is an even one; it should equal to 2, hence $tau_1=1$. All other odd numbers should map to even ones, and vice versa. Hence the number of permutations under the question is the square of the number of bijections from odds (from 3 to $n$) to evens satisfying the same constraint.
If $n$ is even, we cannot have $tau_1=1$, otherwise there would be another even sum. Hence a similar reasoning applies.
It remains to show that there exists at least one such permutation.. (Notice that for the constraint on $i+tau_i$ this would follow easily from the Bertrand postulate...)
answered Nov 15 at 7:58
Ilya Bogdanov
13.5k2858
4
I believe it is still an open problem whether there are infinitely many primes of the form $a^4+b^4$. I can't provide a reference right now, but I have seen it claimed that a Friedlander-Iwaniec-type result regarding primes of the form $a^3+2b^3$ that this is the sparsest polynomial set which is known to contain infinitely many primes.
– Wojowu
Nov 15 at 8:50
9
There is no such permutation if $n=12$. @MTyson's comment above lists $S(12)=0$. (You don't need to check all $12!$ permutations. The numbers $12^4+1^4,12^4+3^4,12^4+5^4,dotsc,12^4+11^4$ are all composite, that's sufficient.)
– Zach Teitler
Nov 15 at 11:11
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
21
down vote
The fact on squareness is easy.
If $n$ is odd, among the sums of the form $i^4+tau_i^4$ there is an even one; it should equal to 2, hence $tau_1=1$. All other odd numbers should map to even ones, and vice versa. Hence the number of permutations under the question is the square of the number of bijections from odds (from 3 to $n$) to evens satisfying the same constraint.
If $n$ is even, we cannot have $tau_1=1$, otherwise there would be another even sum. Hence a similar reasoning applies.
It remains to show that there exists at least one such permutation.. (Notice that for the constraint on $i+tau_i$ this would follow easily from the Bertrand postulate...)
answered Nov 15 at 7:58
Ilya Bogdanov
13.5k2858
4
I believe it is still an open problem whether there are infinitely many primes of the form $a^4+b^4$. I can't provide a reference right now, but I have seen it claimed that a Friedlander-Iwaniec-type result regarding primes of the form $a^3+2b^3$ that this is the sparsest polynomial set which is known to contain infinitely many primes.
– Wojowu
Nov 15 at 8:50
9
There is no such permutation if $n=12$. @MTyson's comment above lists $S(12)=0$. (You don't need to check all $12!$ permutations. The numbers $12^4+1^4,12^4+3^4,12^4+5^4,dotsc,12^4+11^4$ are all composite, that's sufficient.)
– Zach Teitler
Nov 15 at 11:11
add a comment |
up vote
21
down vote
The fact on squareness is easy.
If $n$ is odd, among the sums of the form $i^4+tau_i^4$ there is an even one; it should equal to 2, hence $tau_1=1$. All other odd numbers should map to even ones, and vice versa. Hence the number of permutations under the question is the square of the number of bijections from odds (from 3 to $n$) to evens satisfying the same constraint.
If $n$ is even, we cannot have $tau_1=1$, otherwise there would be another even sum. Hence a similar reasoning applies.
It remains to show that there exists at least one such permutation.. (Notice that for the constraint on $i+tau_i$ this would follow easily from the Bertrand postulate...)
answered Nov 15 at 7:58
Ilya Bogdanov
13.5k2858
4
I believe it is still an open problem whether there are infinitely many primes of the form $a^4+b^4$. I can't provide a reference right now, but I have seen it claimed that a Friedlander-Iwaniec-type result regarding primes of the form $a^3+2b^3$ that this is the sparsest polynomial set which is known to contain infinitely many primes.
– Wojowu
Nov 15 at 8:50
9
There is no such permutation if $n=12$. @MTyson's comment above lists $S(12)=0$. (You don't need to check all $12!$ permutations. The numbers $12^4+1^4,12^4+3^4,12^4+5^4,dotsc,12^4+11^4$ are all composite, that's sufficient.)
– Zach Teitler
Nov 15 at 11:11
add a comment |
up vote
21
down vote
up vote
21
down vote
The fact on squareness is easy.
If $n$ is odd, among the sums of the form $i^4+tau_i^4$ there is an even one; it should equal to 2, hence $tau_1=1$. All other odd numbers should map to even ones, and vice versa. Hence the number of permutations under the question is the square of the number of bijections from odds (from 3 to $n$) to evens satisfying the same constraint.
If $n$ is even, we cannot have $tau_1=1$, otherwise there would be another even sum. Hence a similar reasoning applies.
It remains to show that there exists at least one such permutation.. (Notice that for the constraint on $i+tau_i$ this would follow easily from the Bertrand postulate...)
answered Nov 15 at 7:58
Ilya Bogdanov
13.5k2858
The fact on squareness is easy.
If $n$ is odd, among the sums of the form $i^4+tau_i^4$ there is an even one; it should equal to 2, hence $tau_1=1$. All other odd numbers should map to even ones, and vice versa. Hence the number of permutations under the question is the square of the number of bijections from odds (from 3 to $n$) to evens satisfying the same constraint.
If $n$ is even, we cannot have $tau_1=1$, otherwise there would be another even sum. Hence a similar reasoning applies.
It remains to show that there exists at least one such permutation.. (Notice that for the constraint on $i+tau_i$ this would follow easily from the Bertrand postulate...)
answered Nov 15 at 7:58
Ilya Bogdanov
13.5k2858
answered Nov 15 at 7:58
Ilya Bogdanov
13.5k2858
answered Nov 15 at 7:58
Ilya Bogdanov
13.5k2858
answered Nov 15 at 7:58
Ilya Bogdanov
13.5k2858
13.5k2858
4
I believe it is still an open problem whether there are infinitely many primes of the form $a^4+b^4$. I can't provide a reference right now, but I have seen it claimed that a Friedlander-Iwaniec-type result regarding primes of the form $a^3+2b^3$ that this is the sparsest polynomial set which is known to contain infinitely many primes.
– Wojowu
Nov 15 at 8:50
9
There is no such permutation if $n=12$. @MTyson's comment above lists $S(12)=0$. (You don't need to check all $12!$ permutations. The numbers $12^4+1^4,12^4+3^4,12^4+5^4,dotsc,12^4+11^4$ are all composite, that's sufficient.)
– Zach Teitler
Nov 15 at 11:11
add a comment |
4
I believe it is still an open problem whether there are infinitely many primes of the form $a^4+b^4$. I can't provide a reference right now, but I have seen it claimed that a Friedlander-Iwaniec-type result regarding primes of the form $a^3+2b^3$ that this is the sparsest polynomial set which is known to contain infinitely many primes.
– Wojowu
Nov 15 at 8:50
9
There is no such permutation if $n=12$. @MTyson's comment above lists $S(12)=0$. (You don't need to check all $12!$ permutations. The numbers $12^4+1^4,12^4+3^4,12^4+5^4,dotsc,12^4+11^4$ are all composite, that's sufficient.)
– Zach Teitler
Nov 15 at 11:11
4
4
I believe it is still an open problem whether there are infinitely many primes of the form $a^4+b^4$. I can't provide a reference right now, but I have seen it claimed that a Friedlander-Iwaniec-type result regarding primes of the form $a^3+2b^3$ that this is the sparsest polynomial set which is known to contain infinitely many primes.
– Wojowu
Nov 15 at 8:50
I believe it is still an open problem whether there are infinitely many primes of the form $a^4+b^4$. I can't provide a reference right now, but I have seen it claimed that a Friedlander-Iwaniec-type result regarding primes of the form $a^3+2b^3$ that this is the sparsest polynomial set which is known to contain infinitely many primes.
– Wojowu
Nov 15 at 8:50
9
9
There is no such permutation if $n=12$. @MTyson's comment above lists $S(12)=0$. (You don't need to check all $12!$ permutations. The numbers $12^4+1^4,12^4+3^4,12^4+5^4,dotsc,12^4+11^4$ are all composite, that's sufficient.)
– Zach Teitler
Nov 15 at 11:11
There is no such permutation if $n=12$. @MTyson's comment above lists $S(12)=0$. (You don't need to check all $12!$ permutations. The numbers $12^4+1^4,12^4+3^4,12^4+5^4,dotsc,12^4+11^4$ are all composite, that's sufficient.)
– Zach Teitler
Nov 15 at 11:11
add a comment |
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$(S(12),dots,S(20))=(0,12^2,12^2,17^2,66^2,54^2,150^2,282^2,1374^2)$. I computed the permanent of the matrix $M$ where $M_{ij}$ is $1$ if $i^4+j^4$ is prime and $0$ otherwise.
– MTyson
Nov 15 at 6:02
7
Note that $M$ is symmetric and that all $1$'s are in entries where $i+j$ is odd, except for the $1$ in the upper left. Therefore by sorting rows and columns into the order $(1,3,5,dots,2,4,6,dots)$ it can be transformed into a block matrix $begin{pmatrix}W&X\X^top& 0end{pmatrix}$, where $W$ is $0$ except for a $1$ in the upper left entry. If it weren't for this $1$, we would have $mathrm{perm}(M)=mathrm{perm}(X)^2$.
– MTyson
Nov 15 at 6:29
2
@MTyson it doesn't matter if $W$ has $1$ in the upper left entry or has $0$. $mathrm{perm}(M)=mathrm{perm}(X)^2$ still holds.
– Nemo
Nov 15 at 7:38
1
@Nemo When $n$ is odd, $X$ is not even a square matrix...
– Zach Teitler
Nov 15 at 11:17
1
I think this conjecture was hasty and both @MTyson comments and Ilya Bogdanov's answer show the result has very little to do with primes.
– Yaakov Baruch
Nov 16 at 9:45