What is the general solution of $x+y+z mid xyz$?
Is there any general solution of the equation $x+y+z mid xyz$ in positive integers where $x,y,z$ are pairwise relatively prime? If $n$ is a certain positive integer, is there any general solution to $x+y+z mid nxyz$ where they are pairwise relatively prime?
Edit:
For a given $x,y$ we can use $z = xy-x-y$ to give a family of infinite solutions. In fact, if $z=k-x-y$, then:
$$k mid xcdot ycdot (k-x-y) implies k mid xy(x+y)$$
Thus, the general solution is $(x,y,z) = (x,y,frac{xy(x+y)}{d}-x-y))$ where $d$ is a divisor of $xy(x+y)$.
Similarly, for a given $n$, we have $(x,y,z) = (x,y,frac{nxy(x+y)}{d}-x-y))$ where $d$ is a divisor of $nxy(x+y)$.
For the pairwise relatively prime part, is there any addition possible?
number-theory divisibility
asked Dec 13 '18 at 9:42
HaranHaran
805322
add a comment |
Is there any general solution of the equation $x+y+z mid xyz$ in positive integers where $x,y,z$ are pairwise relatively prime? If $n$ is a certain positive integer, is there any general solution to $x+y+z mid nxyz$ where they are pairwise relatively prime?
Edit:
For a given $x,y$ we can use $z = xy-x-y$ to give a family of infinite solutions. In fact, if $z=k-x-y$, then:
$$k mid xcdot ycdot (k-x-y) implies k mid xy(x+y)$$
Thus, the general solution is $(x,y,z) = (x,y,frac{xy(x+y)}{d}-x-y))$ where $d$ is a divisor of $xy(x+y)$.
Similarly, for a given $n$, we have $(x,y,z) = (x,y,frac{nxy(x+y)}{d}-x-y))$ where $d$ is a divisor of $nxy(x+y)$.
For the pairwise relatively prime part, is there any addition possible?
number-theory divisibility
asked Dec 13 '18 at 9:42
HaranHaran
805322
add a comment |
Is there any general solution of the equation $x+y+z mid xyz$ in positive integers where $x,y,z$ are pairwise relatively prime? If $n$ is a certain positive integer, is there any general solution to $x+y+z mid nxyz$ where they are pairwise relatively prime?
Edit:
For a given $x,y$ we can use $z = xy-x-y$ to give a family of infinite solutions. In fact, if $z=k-x-y$, then:
$$k mid xcdot ycdot (k-x-y) implies k mid xy(x+y)$$
Thus, the general solution is $(x,y,z) = (x,y,frac{xy(x+y)}{d}-x-y))$ where $d$ is a divisor of $xy(x+y)$.
Similarly, for a given $n$, we have $(x,y,z) = (x,y,frac{nxy(x+y)}{d}-x-y))$ where $d$ is a divisor of $nxy(x+y)$.
For the pairwise relatively prime part, is there any addition possible?
number-theory divisibility
asked Dec 13 '18 at 9:42
HaranHaran
805322
Is there any general solution of the equation $x+y+z mid xyz$ in positive integers where $x,y,z$ are pairwise relatively prime? If $n$ is a certain positive integer, is there any general solution to $x+y+z mid nxyz$ where they are pairwise relatively prime?
Edit:
For a given $x,y$ we can use $z = xy-x-y$ to give a family of infinite solutions. In fact, if $z=k-x-y$, then:
$$k mid xcdot ycdot (k-x-y) implies k mid xy(x+y)$$
Thus, the general solution is $(x,y,z) = (x,y,frac{xy(x+y)}{d}-x-y))$ where $d$ is a divisor of $xy(x+y)$.
Similarly, for a given $n$, we have $(x,y,z) = (x,y,frac{nxy(x+y)}{d}-x-y))$ where $d$ is a divisor of $nxy(x+y)$.
For the pairwise relatively prime part, is there any addition possible?
number-theory divisibility
number-theory divisibility
asked Dec 13 '18 at 9:42
HaranHaran
805322
asked Dec 13 '18 at 9:42
HaranHaran
805322
edited Dec 13 '18 at 14:19
Haran
asked Dec 13 '18 at 9:42
HaranHaran
805322
asked Dec 13 '18 at 9:42
HaranHaran
805322
asked Dec 13 '18 at 9:42
HaranHaran
805322
805322
add a comment |
add a comment |
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