Can we solve a simple set of equations in radicals?












2












$begingroup$


Suppose $x,y,z$ are non-zero coprime integers with $x+y ne z$ and which satisfy the following three equations:




  • $operatorname{rad}(x)=operatorname{rad}(z-y)$

  • $operatorname{rad}(y)=operatorname{rad}(z-x)$

  • $operatorname{rad}(z)=operatorname{rad}(x+y)$


Here, the radical of $n$ is defined as the product of all prime divisors of $n$. Hence $operatorname{rad}(n)=prod_{p|n}{p}$.



It is not difficult to check that if $(x,y,z)$ is a solution, also $(-x,-y,-z)$ and $(y,x,z),(z,-x,y),(-y,z,x)$ are solutions. Beside these symmetries, can we prove that $(5,27,2)$ is the only coprime solution?






number-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    Have you checked this for a number of values?
    $endgroup$
    – marty cohen
    Nov 1 '15 at 22:34










  • $begingroup$
    Are these pairwise coprime, or is gcd(x, y, z) = 1?
    $endgroup$
    – marty cohen
    Nov 1 '15 at 22:36










  • $begingroup$
    I checked it for the range $x,y,z in [1, dots ,500]$.
    $endgroup$
    – Rolandb
    Nov 2 '15 at 6:10










  • $begingroup$
    I meant $gcd(x,y,z)=1$ but numerically for this range, pairwise coprime yields the same result. @marty
    $endgroup$
    – Rolandb
    Nov 2 '15 at 6:19
















2












$begingroup$


Suppose $x,y,z$ are non-zero coprime integers with $x+y ne z$ and which satisfy the following three equations:




  • $operatorname{rad}(x)=operatorname{rad}(z-y)$

  • $operatorname{rad}(y)=operatorname{rad}(z-x)$

  • $operatorname{rad}(z)=operatorname{rad}(x+y)$


Here, the radical of $n$ is defined as the product of all prime divisors of $n$. Hence $operatorname{rad}(n)=prod_{p|n}{p}$.



It is not difficult to check that if $(x,y,z)$ is a solution, also $(-x,-y,-z)$ and $(y,x,z),(z,-x,y),(-y,z,x)$ are solutions. Beside these symmetries, can we prove that $(5,27,2)$ is the only coprime solution?






number-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    Have you checked this for a number of values?
    $endgroup$
    – marty cohen
    Nov 1 '15 at 22:34










  • $begingroup$
    Are these pairwise coprime, or is gcd(x, y, z) = 1?
    $endgroup$
    – marty cohen
    Nov 1 '15 at 22:36










  • $begingroup$
    I checked it for the range $x,y,z in [1, dots ,500]$.
    $endgroup$
    – Rolandb
    Nov 2 '15 at 6:10










  • $begingroup$
    I meant $gcd(x,y,z)=1$ but numerically for this range, pairwise coprime yields the same result. @marty
    $endgroup$
    – Rolandb
    Nov 2 '15 at 6:19














2












2








2





$begingroup$


Suppose $x,y,z$ are non-zero coprime integers with $x+y ne z$ and which satisfy the following three equations:




  • $operatorname{rad}(x)=operatorname{rad}(z-y)$

  • $operatorname{rad}(y)=operatorname{rad}(z-x)$

  • $operatorname{rad}(z)=operatorname{rad}(x+y)$


Here, the radical of $n$ is defined as the product of all prime divisors of $n$. Hence $operatorname{rad}(n)=prod_{p|n}{p}$.



It is not difficult to check that if $(x,y,z)$ is a solution, also $(-x,-y,-z)$ and $(y,x,z),(z,-x,y),(-y,z,x)$ are solutions. Beside these symmetries, can we prove that $(5,27,2)$ is the only coprime solution?






number-theory







share|cite|improve this question











$endgroup$




Suppose $x,y,z$ are non-zero coprime integers with $x+y ne z$ and which satisfy the following three equations:




  • $operatorname{rad}(x)=operatorname{rad}(z-y)$

  • $operatorname{rad}(y)=operatorname{rad}(z-x)$

  • $operatorname{rad}(z)=operatorname{rad}(x+y)$


Here, the radical of $n$ is defined as the product of all prime divisors of $n$. Hence $operatorname{rad}(n)=prod_{p|n}{p}$.



It is not difficult to check that if $(x,y,z)$ is a solution, also $(-x,-y,-z)$ and $(y,x,z),(z,-x,y),(-y,z,x)$ are solutions. Beside these symmetries, can we prove that $(5,27,2)$ is the only coprime solution?






number-theory




number-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 22 '18 at 20:43







Rolandb

















asked Nov 1 '15 at 20:56









RolandbRolandb

1226




1226












  • $begingroup$
    Have you checked this for a number of values?
    $endgroup$
    – marty cohen
    Nov 1 '15 at 22:34










  • $begingroup$
    Are these pairwise coprime, or is gcd(x, y, z) = 1?
    $endgroup$
    – marty cohen
    Nov 1 '15 at 22:36










  • $begingroup$
    I checked it for the range $x,y,z in [1, dots ,500]$.
    $endgroup$
    – Rolandb
    Nov 2 '15 at 6:10










  • $begingroup$
    I meant $gcd(x,y,z)=1$ but numerically for this range, pairwise coprime yields the same result. @marty
    $endgroup$
    – Rolandb
    Nov 2 '15 at 6:19


















  • $begingroup$
    Have you checked this for a number of values?
    $endgroup$
    – marty cohen
    Nov 1 '15 at 22:34










  • $begingroup$
    Are these pairwise coprime, or is gcd(x, y, z) = 1?
    $endgroup$
    – marty cohen
    Nov 1 '15 at 22:36










  • $begingroup$
    I checked it for the range $x,y,z in [1, dots ,500]$.
    $endgroup$
    – Rolandb
    Nov 2 '15 at 6:10










  • $begingroup$
    I meant $gcd(x,y,z)=1$ but numerically for this range, pairwise coprime yields the same result. @marty
    $endgroup$
    – Rolandb
    Nov 2 '15 at 6:19
















$begingroup$
Have you checked this for a number of values?
$endgroup$
– marty cohen
Nov 1 '15 at 22:34




$begingroup$
Have you checked this for a number of values?
$endgroup$
– marty cohen
Nov 1 '15 at 22:34












$begingroup$
Are these pairwise coprime, or is gcd(x, y, z) = 1?
$endgroup$
– marty cohen
Nov 1 '15 at 22:36




$begingroup$
Are these pairwise coprime, or is gcd(x, y, z) = 1?
$endgroup$
– marty cohen
Nov 1 '15 at 22:36












$begingroup$
I checked it for the range $x,y,z in [1, dots ,500]$.
$endgroup$
– Rolandb
Nov 2 '15 at 6:10




$begingroup$
I checked it for the range $x,y,z in [1, dots ,500]$.
$endgroup$
– Rolandb
Nov 2 '15 at 6:10












$begingroup$
I meant $gcd(x,y,z)=1$ but numerically for this range, pairwise coprime yields the same result. @marty
$endgroup$
– Rolandb
Nov 2 '15 at 6:19




$begingroup$
I meant $gcd(x,y,z)=1$ but numerically for this range, pairwise coprime yields the same result. @marty
$endgroup$
– Rolandb
Nov 2 '15 at 6:19










1 Answer
1






active

oldest

votes


















-1












$begingroup$

I found a general expression for a solution. For some power $m$ we have an integer $s$ such that $text{rad}(x)^m=sx$. For instance $x=-2^3.3$ we have $text{rad}(x)=2.3$ and $s=-3^2$.
In general $m=n$ works.



Using this observation, there are pairs of non-zero integers $(a,b),(c,d),(e,f)$ such that
$$begin{matrix}
a x=b (z-y)\
c y=d (z-x)\
e z=f (x+y)\
end{matrix}$$
Note that we can assume that the pairs $(a,b),(c,d),(e,f)$ are coprime as we can remove the common factor.



We can solve this set of equations and find
$$x=z.dfrac{b(c-d)}{ac-bd} text{ and } y = z.dfrac{d(a-b)}{ac-bd}$$
We have $ac-bd ne 0$ as $ac=bd$ implies that $xy=(z-y)(z-x)=z(z-x-y)+xy$ and hence $z(z-x-y)=0$ which is not possible.



Clearly, if $x$ and $z$ are coprime we must have that $z mid ac-bd$. How can we prove that $(5,27,2)$ is the only coprime solution (up to symmetry)?






share|cite|improve this answer











$endgroup$













  • $begingroup$
    It's not clear to me what values you mean to give to $x$, $y$, and $z$.
    $endgroup$
    – Gerry Myerson
    Nov 3 '15 at 6:29










  • $begingroup$
    $x=2^{13}$, $y=5^2.7^3$, $z=3^6.23$ and $x=2^7$, $y=5^6.7$, $z=3^2.23^3$.
    $endgroup$
    – Rolandb
    Nov 3 '15 at 19:08










  • $begingroup$
    But the question didn't ask for two sets $x,y,z$ with the same radical of $xyz$. I don't see how what you have written relates to the question at all.
    $endgroup$
    – Gerry Myerson
    Nov 3 '15 at 22:23










  • $begingroup$
    Are you still there?
    $endgroup$
    – Gerry Myerson
    Nov 5 '15 at 4:53










  • $begingroup$
    Apologies. I made a mistake but I could not respond earlier. A proof is added.
    $endgroup$
    – Rolandb
    Nov 8 '15 at 12:22











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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









-1












$begingroup$

I found a general expression for a solution. For some power $m$ we have an integer $s$ such that $text{rad}(x)^m=sx$. For instance $x=-2^3.3$ we have $text{rad}(x)=2.3$ and $s=-3^2$.
In general $m=n$ works.



Using this observation, there are pairs of non-zero integers $(a,b),(c,d),(e,f)$ such that
$$begin{matrix}
a x=b (z-y)\
c y=d (z-x)\
e z=f (x+y)\
end{matrix}$$
Note that we can assume that the pairs $(a,b),(c,d),(e,f)$ are coprime as we can remove the common factor.



We can solve this set of equations and find
$$x=z.dfrac{b(c-d)}{ac-bd} text{ and } y = z.dfrac{d(a-b)}{ac-bd}$$
We have $ac-bd ne 0$ as $ac=bd$ implies that $xy=(z-y)(z-x)=z(z-x-y)+xy$ and hence $z(z-x-y)=0$ which is not possible.



Clearly, if $x$ and $z$ are coprime we must have that $z mid ac-bd$. How can we prove that $(5,27,2)$ is the only coprime solution (up to symmetry)?






share|cite|improve this answer











$endgroup$













  • $begingroup$
    It's not clear to me what values you mean to give to $x$, $y$, and $z$.
    $endgroup$
    – Gerry Myerson
    Nov 3 '15 at 6:29










  • $begingroup$
    $x=2^{13}$, $y=5^2.7^3$, $z=3^6.23$ and $x=2^7$, $y=5^6.7$, $z=3^2.23^3$.
    $endgroup$
    – Rolandb
    Nov 3 '15 at 19:08










  • $begingroup$
    But the question didn't ask for two sets $x,y,z$ with the same radical of $xyz$. I don't see how what you have written relates to the question at all.
    $endgroup$
    – Gerry Myerson
    Nov 3 '15 at 22:23










  • $begingroup$
    Are you still there?
    $endgroup$
    – Gerry Myerson
    Nov 5 '15 at 4:53










  • $begingroup$
    Apologies. I made a mistake but I could not respond earlier. A proof is added.
    $endgroup$
    – Rolandb
    Nov 8 '15 at 12:22
















-1












$begingroup$

I found a general expression for a solution. For some power $m$ we have an integer $s$ such that $text{rad}(x)^m=sx$. For instance $x=-2^3.3$ we have $text{rad}(x)=2.3$ and $s=-3^2$.
In general $m=n$ works.



Using this observation, there are pairs of non-zero integers $(a,b),(c,d),(e,f)$ such that
$$begin{matrix}
a x=b (z-y)\
c y=d (z-x)\
e z=f (x+y)\
end{matrix}$$
Note that we can assume that the pairs $(a,b),(c,d),(e,f)$ are coprime as we can remove the common factor.



We can solve this set of equations and find
$$x=z.dfrac{b(c-d)}{ac-bd} text{ and } y = z.dfrac{d(a-b)}{ac-bd}$$
We have $ac-bd ne 0$ as $ac=bd$ implies that $xy=(z-y)(z-x)=z(z-x-y)+xy$ and hence $z(z-x-y)=0$ which is not possible.



Clearly, if $x$ and $z$ are coprime we must have that $z mid ac-bd$. How can we prove that $(5,27,2)$ is the only coprime solution (up to symmetry)?






share|cite|improve this answer











$endgroup$













  • $begingroup$
    It's not clear to me what values you mean to give to $x$, $y$, and $z$.
    $endgroup$
    – Gerry Myerson
    Nov 3 '15 at 6:29










  • $begingroup$
    $x=2^{13}$, $y=5^2.7^3$, $z=3^6.23$ and $x=2^7$, $y=5^6.7$, $z=3^2.23^3$.
    $endgroup$
    – Rolandb
    Nov 3 '15 at 19:08










  • $begingroup$
    But the question didn't ask for two sets $x,y,z$ with the same radical of $xyz$. I don't see how what you have written relates to the question at all.
    $endgroup$
    – Gerry Myerson
    Nov 3 '15 at 22:23










  • $begingroup$
    Are you still there?
    $endgroup$
    – Gerry Myerson
    Nov 5 '15 at 4:53










  • $begingroup$
    Apologies. I made a mistake but I could not respond earlier. A proof is added.
    $endgroup$
    – Rolandb
    Nov 8 '15 at 12:22














-1












-1








-1





$begingroup$

I found a general expression for a solution. For some power $m$ we have an integer $s$ such that $text{rad}(x)^m=sx$. For instance $x=-2^3.3$ we have $text{rad}(x)=2.3$ and $s=-3^2$.
In general $m=n$ works.



Using this observation, there are pairs of non-zero integers $(a,b),(c,d),(e,f)$ such that
$$begin{matrix}
a x=b (z-y)\
c y=d (z-x)\
e z=f (x+y)\
end{matrix}$$
Note that we can assume that the pairs $(a,b),(c,d),(e,f)$ are coprime as we can remove the common factor.



We can solve this set of equations and find
$$x=z.dfrac{b(c-d)}{ac-bd} text{ and } y = z.dfrac{d(a-b)}{ac-bd}$$
We have $ac-bd ne 0$ as $ac=bd$ implies that $xy=(z-y)(z-x)=z(z-x-y)+xy$ and hence $z(z-x-y)=0$ which is not possible.



Clearly, if $x$ and $z$ are coprime we must have that $z mid ac-bd$. How can we prove that $(5,27,2)$ is the only coprime solution (up to symmetry)?






share|cite|improve this answer











$endgroup$



I found a general expression for a solution. For some power $m$ we have an integer $s$ such that $text{rad}(x)^m=sx$. For instance $x=-2^3.3$ we have $text{rad}(x)=2.3$ and $s=-3^2$.
In general $m=n$ works.



Using this observation, there are pairs of non-zero integers $(a,b),(c,d),(e,f)$ such that
$$begin{matrix}
a x=b (z-y)\
c y=d (z-x)\
e z=f (x+y)\
end{matrix}$$
Note that we can assume that the pairs $(a,b),(c,d),(e,f)$ are coprime as we can remove the common factor.



We can solve this set of equations and find
$$x=z.dfrac{b(c-d)}{ac-bd} text{ and } y = z.dfrac{d(a-b)}{ac-bd}$$
We have $ac-bd ne 0$ as $ac=bd$ implies that $xy=(z-y)(z-x)=z(z-x-y)+xy$ and hence $z(z-x-y)=0$ which is not possible.



Clearly, if $x$ and $z$ are coprime we must have that $z mid ac-bd$. How can we prove that $(5,27,2)$ is the only coprime solution (up to symmetry)?







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 22 '18 at 20:42

























answered Nov 3 '15 at 5:45









RolandbRolandb

1226




1226












  • $begingroup$
    It's not clear to me what values you mean to give to $x$, $y$, and $z$.
    $endgroup$
    – Gerry Myerson
    Nov 3 '15 at 6:29










  • $begingroup$
    $x=2^{13}$, $y=5^2.7^3$, $z=3^6.23$ and $x=2^7$, $y=5^6.7$, $z=3^2.23^3$.
    $endgroup$
    – Rolandb
    Nov 3 '15 at 19:08










  • $begingroup$
    But the question didn't ask for two sets $x,y,z$ with the same radical of $xyz$. I don't see how what you have written relates to the question at all.
    $endgroup$
    – Gerry Myerson
    Nov 3 '15 at 22:23










  • $begingroup$
    Are you still there?
    $endgroup$
    – Gerry Myerson
    Nov 5 '15 at 4:53










  • $begingroup$
    Apologies. I made a mistake but I could not respond earlier. A proof is added.
    $endgroup$
    – Rolandb
    Nov 8 '15 at 12:22


















  • $begingroup$
    It's not clear to me what values you mean to give to $x$, $y$, and $z$.
    $endgroup$
    – Gerry Myerson
    Nov 3 '15 at 6:29










  • $begingroup$
    $x=2^{13}$, $y=5^2.7^3$, $z=3^6.23$ and $x=2^7$, $y=5^6.7$, $z=3^2.23^3$.
    $endgroup$
    – Rolandb
    Nov 3 '15 at 19:08










  • $begingroup$
    But the question didn't ask for two sets $x,y,z$ with the same radical of $xyz$. I don't see how what you have written relates to the question at all.
    $endgroup$
    – Gerry Myerson
    Nov 3 '15 at 22:23










  • $begingroup$
    Are you still there?
    $endgroup$
    – Gerry Myerson
    Nov 5 '15 at 4:53










  • $begingroup$
    Apologies. I made a mistake but I could not respond earlier. A proof is added.
    $endgroup$
    – Rolandb
    Nov 8 '15 at 12:22
















$begingroup$
It's not clear to me what values you mean to give to $x$, $y$, and $z$.
$endgroup$
– Gerry Myerson
Nov 3 '15 at 6:29




$begingroup$
It's not clear to me what values you mean to give to $x$, $y$, and $z$.
$endgroup$
– Gerry Myerson
Nov 3 '15 at 6:29












$begingroup$
$x=2^{13}$, $y=5^2.7^3$, $z=3^6.23$ and $x=2^7$, $y=5^6.7$, $z=3^2.23^3$.
$endgroup$
– Rolandb
Nov 3 '15 at 19:08




$begingroup$
$x=2^{13}$, $y=5^2.7^3$, $z=3^6.23$ and $x=2^7$, $y=5^6.7$, $z=3^2.23^3$.
$endgroup$
– Rolandb
Nov 3 '15 at 19:08












$begingroup$
But the question didn't ask for two sets $x,y,z$ with the same radical of $xyz$. I don't see how what you have written relates to the question at all.
$endgroup$
– Gerry Myerson
Nov 3 '15 at 22:23




$begingroup$
But the question didn't ask for two sets $x,y,z$ with the same radical of $xyz$. I don't see how what you have written relates to the question at all.
$endgroup$
– Gerry Myerson
Nov 3 '15 at 22:23












$begingroup$
Are you still there?
$endgroup$
– Gerry Myerson
Nov 5 '15 at 4:53




$begingroup$
Are you still there?
$endgroup$
– Gerry Myerson
Nov 5 '15 at 4:53












$begingroup$
Apologies. I made a mistake but I could not respond earlier. A proof is added.
$endgroup$
– Rolandb
Nov 8 '15 at 12:22




$begingroup$
Apologies. I made a mistake but I could not respond earlier. A proof is added.
$endgroup$
– Rolandb
Nov 8 '15 at 12:22


















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Marschland

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Kilometerweite Marsch in den Vier- und Marschlanden am Elbdeich Die Wedeler Marschlandschaft in Schleswig-Holstein Der Hadelner Kanal zur Entwässerung der Marsch Die tief gelegene Marsch und ein Entwässerungskanal Das Schöpfwerk Otterndorf im Land Hadeln zur Entwässerung der Medem verfügte damals bereits über die größte Kreiselpumpe Europas. Das Schöpfwerk in Neuhaus für die Aue und des Neuhaus-Bülkauer Kanals Als Marsch(land) (v. niederdt., altsächs. mersc ) – auch Masch , Mersch oder Schwemmland genannt – bezeichnet man eine nacheiszeitlich entstandene geomorphologische Landform im Gebiet der nordwestdeutschen Küsten und Flüsse sowie vergleichbare Landformen weltweit. Inhaltsverzeichnis 1 Landschaft 2 Entstehung 3 Marschböden 4 Nutzung 5 Marschgebiete in Deutschland 6 Siehe auch 7 Literatur 8 Weblinks 9 Einzelnachweise Landschaft | Marschen sind generell flache Landstriche ohne natürliche Erhebungen. Si
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Dieringhausen

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Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
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