Can we solve a simple set of equations in radicals?
$begingroup$
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
asked Nov 1 '15 at 20:56
RolandbRolandb
1226
$endgroup$
add a comment |
$begingroup$
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
asked Nov 1 '15 at 20:56
RolandbRolandb
1226
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
asked Nov 1 '15 at 20:56
RolandbRolandb
1226
$endgroup$
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
asked Nov 1 '15 at 20:56
RolandbRolandb
1226
asked Nov 1 '15 at 20:56
RolandbRolandb
1226
edited Dec 22 '18 at 20:43
Rolandb
asked Nov 1 '15 at 20:56
RolandbRolandb
1226
asked Nov 1 '15 at 20:56
RolandbRolandb
1226
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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)?
answered Nov 3 '15 at 5:45
RolandbRolandb
1226
$endgroup$
$begingroup$
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
|
show 6 more comments
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1 Answer
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1 Answer
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oldest
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$begingroup$
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)?
answered Nov 3 '15 at 5:45
RolandbRolandb
1226
$endgroup$
$begingroup$
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
|
show 6 more comments
$begingroup$
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)?
answered Nov 3 '15 at 5:45
RolandbRolandb
1226
$endgroup$
$begingroup$
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
|
show 6 more comments
$begingroup$
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)?
answered Nov 3 '15 at 5:45
RolandbRolandb
1226
$endgroup$
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)?
answered Nov 3 '15 at 5:45
RolandbRolandb
1226
edited Dec 22 '18 at 20:42
answered Nov 3 '15 at 5:45
RolandbRolandb
1226
answered Nov 3 '15 at 5:45
RolandbRolandb
1226
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
|
show 6 more comments
$begingroup$
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
|
show 6 more comments
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$begingroup$
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