A mysterious connection between primes and squares
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16
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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 21 hours ago
Zhi-Wei Sun
2,041222
add a comment |
up vote
16
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 21 hours ago
Zhi-Wei Sun
2,041222
Any further check of this conjecture is welcome! By the way, it might be possible to prove that $S(n)$ is always a square.
– Zhi-Wei Sun
21 hours ago
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
20 hours ago
6
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
19 hours ago
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
18 hours ago
1
@Nemo When $n$ is odd, $X$ is not even a square matrix...
– Zach Teitler
15 hours ago
add a comment |
up vote
16
down vote
favorite
up vote
16
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 21 hours ago
Zhi-Wei Sun
2,041222
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 21 hours ago
Zhi-Wei Sun
2,041222
asked 21 hours ago
Zhi-Wei Sun
2,041222
asked 21 hours ago
Zhi-Wei Sun
2,041222
asked 21 hours ago
Zhi-Wei Sun
2,041222
asked 21 hours ago
Zhi-Wei Sun
2,041222
2,041222
Any further check of this conjecture is welcome! By the way, it might be possible to prove that $S(n)$ is always a square.
– Zhi-Wei Sun
21 hours ago
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
20 hours ago
6
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
19 hours ago
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
18 hours ago
1
@Nemo When $n$ is odd, $X$ is not even a square matrix...
– Zach Teitler
15 hours ago
add a comment |
Any further check of this conjecture is welcome! By the way, it might be possible to prove that $S(n)$ is always a square.
– Zhi-Wei Sun
21 hours ago
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
20 hours ago
6
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
19 hours ago
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
18 hours ago
1
@Nemo When $n$ is odd, $X$ is not even a square matrix...
– Zach Teitler
15 hours ago
Any further check of this conjecture is welcome! By the way, it might be possible to prove that $S(n)$ is always a square.
– Zhi-Wei Sun
21 hours ago
Any further check of this conjecture is welcome! By the way, it might be possible to prove that $S(n)$ is always a square.
– Zhi-Wei Sun
21 hours ago
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
20 hours ago
$(S(12),dots,S(20))=(0,12^2,12^2,17^2,66^2,54^2,150^2,282^2,1374^2)$. I computed the permanent of the matrix $M$ where $M_{ij}$ is $1$ if $i^4+j^4$ is prime and $0$ otherwise.
– MTyson
20 hours ago
6
6
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
19 hours ago
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
19 hours ago
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
18 hours ago
@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
18 hours ago
1
1
@Nemo When $n$ is odd, $X$ is not even a square matrix...
– Zach Teitler
15 hours ago
@Nemo When $n$ is odd, $X$ is not even a square matrix...
– Zach Teitler
15 hours ago
add a comment |
1 Answer
1
active
oldest
votes
up vote
19
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 18 hours ago
Ilya Bogdanov
13.4k2858
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
17 hours ago
7
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
15 hours ago
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
19
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 18 hours ago
Ilya Bogdanov
13.4k2858
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
17 hours ago
7
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
15 hours ago
add a comment |
up vote
19
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 18 hours ago
Ilya Bogdanov
13.4k2858
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
17 hours ago
7
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
15 hours ago
add a comment |
up vote
19
down vote
up vote
19
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 18 hours ago
Ilya Bogdanov
13.4k2858
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 18 hours ago
Ilya Bogdanov
13.4k2858
answered 18 hours ago
Ilya Bogdanov
13.4k2858
answered 18 hours ago
Ilya Bogdanov
13.4k2858
answered 18 hours ago
Ilya Bogdanov
13.4k2858
13.4k2858
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
17 hours ago
7
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
15 hours ago
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
17 hours ago
7
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
15 hours ago
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
17 hours ago
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
17 hours ago
7
7
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
15 hours ago
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
15 hours ago
add a comment |
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Any further check of this conjecture is welcome! By the way, it might be possible to prove that $S(n)$ is always a square.
– Zhi-Wei Sun
21 hours ago
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
20 hours ago
6
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
19 hours ago
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
18 hours ago
1
@Nemo When $n$ is odd, $X$ is not even a square matrix...
– Zach Teitler
15 hours ago