A mysterious connection between primes and squares












17












$begingroup$


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







share|cite|improve this question









$endgroup$








  • 10




    $begingroup$
    $(S(12),dots,S(20))=(0,12^2,12^2,17^2,66^2,54^2,150^2,282^2,1374^2)$. I computed the permanent of the matrix $M$ where $M_{ij}$ is $1$ if $i^4+j^4$ is prime and $0$ otherwise.
    $endgroup$
    – MTyson
    Nov 15 '18 at 6:02






  • 7




    $begingroup$
    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 $mathrm{perm}(M)=mathrm{perm}(X)^2$.
    $endgroup$
    – MTyson
    Nov 15 '18 at 6:29






  • 2




    $begingroup$
    @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.
    $endgroup$
    – Nemo
    Nov 15 '18 at 7:38








  • 1




    $begingroup$
    @Nemo When $n$ is odd, $X$ is not even a square matrix...
    $endgroup$
    – Zach Teitler
    Nov 15 '18 at 11:17






  • 1




    $begingroup$
    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.
    $endgroup$
    – Yaakov Baruch
    Nov 16 '18 at 9:45
















17












$begingroup$


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







share|cite|improve this question









$endgroup$








  • 10




    $begingroup$
    $(S(12),dots,S(20))=(0,12^2,12^2,17^2,66^2,54^2,150^2,282^2,1374^2)$. I computed the permanent of the matrix $M$ where $M_{ij}$ is $1$ if $i^4+j^4$ is prime and $0$ otherwise.
    $endgroup$
    – MTyson
    Nov 15 '18 at 6:02






  • 7




    $begingroup$
    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 $mathrm{perm}(M)=mathrm{perm}(X)^2$.
    $endgroup$
    – MTyson
    Nov 15 '18 at 6:29






  • 2




    $begingroup$
    @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.
    $endgroup$
    – Nemo
    Nov 15 '18 at 7:38








  • 1




    $begingroup$
    @Nemo When $n$ is odd, $X$ is not even a square matrix...
    $endgroup$
    – Zach Teitler
    Nov 15 '18 at 11:17






  • 1




    $begingroup$
    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.
    $endgroup$
    – Yaakov Baruch
    Nov 16 '18 at 9:45














17












17








17


5



$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 15 '18 at 4:36









Zhi-Wei SunZhi-Wei Sun

3,327327




3,327327








  • 10




    $begingroup$
    $(S(12),dots,S(20))=(0,12^2,12^2,17^2,66^2,54^2,150^2,282^2,1374^2)$. I computed the permanent of the matrix $M$ where $M_{ij}$ is $1$ if $i^4+j^4$ is prime and $0$ otherwise.
    $endgroup$
    – MTyson
    Nov 15 '18 at 6:02






  • 7




    $begingroup$
    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 $mathrm{perm}(M)=mathrm{perm}(X)^2$.
    $endgroup$
    – MTyson
    Nov 15 '18 at 6:29






  • 2




    $begingroup$
    @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.
    $endgroup$
    – Nemo
    Nov 15 '18 at 7:38








  • 1




    $begingroup$
    @Nemo When $n$ is odd, $X$ is not even a square matrix...
    $endgroup$
    – Zach Teitler
    Nov 15 '18 at 11:17






  • 1




    $begingroup$
    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.
    $endgroup$
    – Yaakov Baruch
    Nov 16 '18 at 9:45














  • 10




    $begingroup$
    $(S(12),dots,S(20))=(0,12^2,12^2,17^2,66^2,54^2,150^2,282^2,1374^2)$. I computed the permanent of the matrix $M$ where $M_{ij}$ is $1$ if $i^4+j^4$ is prime and $0$ otherwise.
    $endgroup$
    – MTyson
    Nov 15 '18 at 6:02






  • 7




    $begingroup$
    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 $mathrm{perm}(M)=mathrm{perm}(X)^2$.
    $endgroup$
    – MTyson
    Nov 15 '18 at 6:29






  • 2




    $begingroup$
    @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.
    $endgroup$
    – Nemo
    Nov 15 '18 at 7:38








  • 1




    $begingroup$
    @Nemo When $n$ is odd, $X$ is not even a square matrix...
    $endgroup$
    – Zach Teitler
    Nov 15 '18 at 11:17






  • 1




    $begingroup$
    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.
    $endgroup$
    – Yaakov Baruch
    Nov 16 '18 at 9:45








10




10




$begingroup$
$(S(12),dots,S(20))=(0,12^2,12^2,17^2,66^2,54^2,150^2,282^2,1374^2)$. I computed the permanent of the matrix $M$ where $M_{ij}$ is $1$ if $i^4+j^4$ is prime and $0$ otherwise.
$endgroup$
– MTyson
Nov 15 '18 at 6:02




$begingroup$
$(S(12),dots,S(20))=(0,12^2,12^2,17^2,66^2,54^2,150^2,282^2,1374^2)$. I computed the permanent of the matrix $M$ where $M_{ij}$ is $1$ if $i^4+j^4$ is prime and $0$ otherwise.
$endgroup$
– MTyson
Nov 15 '18 at 6:02




7




7




$begingroup$
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 $mathrm{perm}(M)=mathrm{perm}(X)^2$.
$endgroup$
– MTyson
Nov 15 '18 at 6:29




$begingroup$
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 $mathrm{perm}(M)=mathrm{perm}(X)^2$.
$endgroup$
– MTyson
Nov 15 '18 at 6:29




2




2




$begingroup$
@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.
$endgroup$
– Nemo
Nov 15 '18 at 7:38






$begingroup$
@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.
$endgroup$
– Nemo
Nov 15 '18 at 7:38






1




1




$begingroup$
@Nemo When $n$ is odd, $X$ is not even a square matrix...
$endgroup$
– Zach Teitler
Nov 15 '18 at 11:17




$begingroup$
@Nemo When $n$ is odd, $X$ is not even a square matrix...
$endgroup$
– Zach Teitler
Nov 15 '18 at 11:17




1




1




$begingroup$
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.
$endgroup$
– Yaakov Baruch
Nov 16 '18 at 9:45




$begingroup$
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.
$endgroup$
– Yaakov Baruch
Nov 16 '18 at 9:45










1 Answer
1






active

oldest

votes


















23












$begingroup$

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...)






share|cite|improve this answer









$endgroup$









  • 4




    $begingroup$
    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.
    $endgroup$
    – Wojowu
    Nov 15 '18 at 8:50








  • 9




    $begingroup$
    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.)
    $endgroup$
    – Zach Teitler
    Nov 15 '18 at 11:11











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1 Answer
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1 Answer
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active

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23












$begingroup$

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...)






share|cite|improve this answer









$endgroup$









  • 4




    $begingroup$
    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.
    $endgroup$
    – Wojowu
    Nov 15 '18 at 8:50








  • 9




    $begingroup$
    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.)
    $endgroup$
    – Zach Teitler
    Nov 15 '18 at 11:11
















23












$begingroup$

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...)






share|cite|improve this answer









$endgroup$









  • 4




    $begingroup$
    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.
    $endgroup$
    – Wojowu
    Nov 15 '18 at 8:50








  • 9




    $begingroup$
    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.)
    $endgroup$
    – Zach Teitler
    Nov 15 '18 at 11:11














23












23








23





$begingroup$

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...)






share|cite|improve this answer









$endgroup$



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...)







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 15 '18 at 7:58









Ilya BogdanovIlya Bogdanov

13.7k2858




13.7k2858








  • 4




    $begingroup$
    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.
    $endgroup$
    – Wojowu
    Nov 15 '18 at 8:50








  • 9




    $begingroup$
    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.)
    $endgroup$
    – Zach Teitler
    Nov 15 '18 at 11:11














  • 4




    $begingroup$
    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.
    $endgroup$
    – Wojowu
    Nov 15 '18 at 8:50








  • 9




    $begingroup$
    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.)
    $endgroup$
    – Zach Teitler
    Nov 15 '18 at 11:11








4




4




$begingroup$
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.
$endgroup$
– Wojowu
Nov 15 '18 at 8:50






$begingroup$
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.
$endgroup$
– Wojowu
Nov 15 '18 at 8:50






9




9




$begingroup$
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.)
$endgroup$
– Zach Teitler
Nov 15 '18 at 11:11




$begingroup$
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.)
$endgroup$
– Zach Teitler
Nov 15 '18 at 11:11


















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