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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Bressuire

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Subprefecture and commune in Nouvelle-Aquitaine, France Bressuire Subprefecture and commune Chateau de Bressuire and the Eglise Notre-Dame Coat of arms Location of Bressuire Bressuire Show map of France Bressuire Show map of Nouvelle-Aquitaine Coordinates: 46°50′27″N 0°29′14″W  /  46.8408°N 0.4872°W  / 46.8408; -0.4872 Coordinates: 46°50′27″N 0°29′14″W  /  46.8408°N 0.4872°W  / 46.8408; -0.4872 Country France Region Nouvelle-Aquitaine Department Deux-Sèvres Arrondissement Bressuire Canton Bressuire Government  • Mayor .mw-parser-output .nobold{font-weight:normal} (2014–20) Jean Michel Bernier Area 1 180.59 km 2 (69.73 sq mi) Population (2014) 2 19,300  • Density 110/km 2 (280/sq mi) Time zone UTC+01:00 (CET)  • Summer (DST) UTC+02:00 (CEST) INSEE/Postal code 79049 /79300 Elevation 98–236 m (322–774 ft) (avg. 173 m or 568 ft) 1 French Land Register data, which exclude...
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Vorschmack

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Vorschmack Ukrainian Jewish-style vorschmack served on rye bread Course Hors d'oeuvre Region or state Eastern Europe Associated national cuisine Ashkenazi Jewish, Finnish, German, Ukrainian, Polish, Russian Main ingredients Ground meat and/or fish Cookbook: Vorschmack   Media: Vorschmack Vorschmack or forshmak (Yiddish: פֿאָרשמאַק ‎, from archaic German Vorschmack , "foretaste" [1] or "appetizer" [2] ) is an originally East European dish made of salty minced fish or meat. Different variants of this dish are especially common in Ashkenazi Jewish and Finnish cuisine. Some varieties are also known in Russian and Polish cuisine. Contents 1 In Jewish cuisine 2 In Russian cuisine 3 In Polish cuisine 4 In Finnish cuisine 5 See also 6 References In Jewish cuisine According to Gil Marks, the German name points to the possible Germanic origin of this dish. [1] William Pokhlyobkin descr...
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Quarantine

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For other uses, see Quarantine (disambiguation). Signal flag "Lima" called the "Yellow Jack" which when flown in harbor means ship is under quarantine. A simple yellow flag (also called the "Yellow Jack") had historically been used to signal quarantine (it stands for Q among signal flags), but now indicates the opposite, as a signal of a ship free of disease that requests boarding and inspection. A quarantine is used to separate and restrict the movement of people; it is 'a restraint upon the activities or communication of persons or the transport of goods designed to prevent the spread of disease or pests', for a certain period of time. [1] This is often used in connection to disease and illness, such as those who may possibly have been exposed to a communicable disease. [2] The term is often erroneously used to mean medical isolation, which is "to separate ill persons who have a communicable disease from those who are healthy...
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