Existence of only finitely many solutions
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Let $r$ be a rational number greater than $1$. Prove that there are only finitely many natural numbers $x,y,z$ such that $$(x+1)(y+1)(z+1)=rxyz.$$
Progress: For $r=8$, the only solutions are $x=y=z=1$. If $r>8$, then clearly there are no solutions. I'm having trouble showing the same for $rin(1,8)$. Any hints or solutions are welcome.
algebra-precalculus elementary-number-theory
asked Nov 11 at 11:44
MathIsFun.
20318
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Let $r$ be a rational number greater than $1$. Prove that there are only finitely many natural numbers $x,y,z$ such that $$(x+1)(y+1)(z+1)=rxyz.$$
Progress: For $r=8$, the only solutions are $x=y=z=1$. If $r>8$, then clearly there are no solutions. I'm having trouble showing the same for $rin(1,8)$. Any hints or solutions are welcome.
algebra-precalculus elementary-number-theory
asked Nov 11 at 11:44
MathIsFun.
20318
*For $ r=0 $?
– Raptor
Nov 11 at 11:54
1
@Raptor rational number greater than 1
– KGSH bteam Mine Team Beast O_
Nov 11 at 12:09
Show m(x + 1)(y + 1)(z + 1) = nxyz has finitely many x,y,z solutions for natural numbers m,n,x,y,z.
– William Elliot
Nov 11 at 22:16
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $r$ be a rational number greater than $1$. Prove that there are only finitely many natural numbers $x,y,z$ such that $$(x+1)(y+1)(z+1)=rxyz.$$
Progress: For $r=8$, the only solutions are $x=y=z=1$. If $r>8$, then clearly there are no solutions. I'm having trouble showing the same for $rin(1,8)$. Any hints or solutions are welcome.
algebra-precalculus elementary-number-theory
asked Nov 11 at 11:44
MathIsFun.
20318
Let $r$ be a rational number greater than $1$. Prove that there are only finitely many natural numbers $x,y,z$ such that $$(x+1)(y+1)(z+1)=rxyz.$$
Progress: For $r=8$, the only solutions are $x=y=z=1$. If $r>8$, then clearly there are no solutions. I'm having trouble showing the same for $rin(1,8)$. Any hints or solutions are welcome.
algebra-precalculus elementary-number-theory
algebra-precalculus elementary-number-theory
asked Nov 11 at 11:44
MathIsFun.
20318
asked Nov 11 at 11:44
MathIsFun.
20318
edited Nov 11 at 12:37
asked Nov 11 at 11:44
MathIsFun.
20318
asked Nov 11 at 11:44
MathIsFun.
20318
asked Nov 11 at 11:44
MathIsFun.
20318
20318
*For $ r=0 $?
– Raptor
Nov 11 at 11:54
1
@Raptor rational number greater than 1
– KGSH bteam Mine Team Beast O_
Nov 11 at 12:09
Show m(x + 1)(y + 1)(z + 1) = nxyz has finitely many x,y,z solutions for natural numbers m,n,x,y,z.
– William Elliot
Nov 11 at 22:16
add a comment |
*For $ r=0 $?
– Raptor
Nov 11 at 11:54
1
@Raptor rational number greater than 1
– KGSH bteam Mine Team Beast O_
Nov 11 at 12:09
Show m(x + 1)(y + 1)(z + 1) = nxyz has finitely many x,y,z solutions for natural numbers m,n,x,y,z.
– William Elliot
Nov 11 at 22:16
*For $ r=0 $?
– Raptor
Nov 11 at 11:54
*For $ r=0 $?
– Raptor
Nov 11 at 11:54
1
1
@Raptor rational number greater than 1
– KGSH bteam Mine Team Beast O_
Nov 11 at 12:09
@Raptor rational number greater than 1
– KGSH bteam Mine Team Beast O_
Nov 11 at 12:09
Show m(x + 1)(y + 1)(z + 1) = nxyz has finitely many x,y,z solutions for natural numbers m,n,x,y,z.
– William Elliot
Nov 11 at 22:16
Show m(x + 1)(y + 1)(z + 1) = nxyz has finitely many x,y,z solutions for natural numbers m,n,x,y,z.
– William Elliot
Nov 11 at 22:16
add a comment |
1 Answer
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x,y,z positive integers.
(1 + 1/x)(1 + 1/y)(1 + 1/z) = r.
Often there will be no solution. r = 7 for example.
Assume that x,y and z's of the solutions diverge.
That forces r = 1, contradicting the assumption of solutions.
Wlog assume the z's are bounded and the x,y's of solutions diverge.
For each solution, there's another solution with larger x,y.
Thus (1 + 1/x)(1 + 1/y) is smaller, 1 + 1/z is larger and z is
smaller. But z can be smaller only finite many times.
Wlog, assume y,z's of the solutions are bounded and x's diverge.
Each time x gets larger, 1 + 1/x gets smaller and (1 + 1/y)(1 + 1/z)
gets larger but only finite many times because each time x,y change
and there's just finite many ways they can change.
Thus the x,y,z's of solutions are all bounded
resulting in only finite many possible solutions.
answered Nov 16 at 9:10
William Elliot
6,7802518
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
x,y,z positive integers.
(1 + 1/x)(1 + 1/y)(1 + 1/z) = r.
Often there will be no solution. r = 7 for example.
Assume that x,y and z's of the solutions diverge.
That forces r = 1, contradicting the assumption of solutions.
Wlog assume the z's are bounded and the x,y's of solutions diverge.
For each solution, there's another solution with larger x,y.
Thus (1 + 1/x)(1 + 1/y) is smaller, 1 + 1/z is larger and z is
smaller. But z can be smaller only finite many times.
Wlog, assume y,z's of the solutions are bounded and x's diverge.
Each time x gets larger, 1 + 1/x gets smaller and (1 + 1/y)(1 + 1/z)
gets larger but only finite many times because each time x,y change
and there's just finite many ways they can change.
Thus the x,y,z's of solutions are all bounded
resulting in only finite many possible solutions.
answered Nov 16 at 9:10
William Elliot
6,7802518
add a comment |
up vote
0
down vote
x,y,z positive integers.
(1 + 1/x)(1 + 1/y)(1 + 1/z) = r.
Often there will be no solution. r = 7 for example.
Assume that x,y and z's of the solutions diverge.
That forces r = 1, contradicting the assumption of solutions.
Wlog assume the z's are bounded and the x,y's of solutions diverge.
For each solution, there's another solution with larger x,y.
Thus (1 + 1/x)(1 + 1/y) is smaller, 1 + 1/z is larger and z is
smaller. But z can be smaller only finite many times.
Wlog, assume y,z's of the solutions are bounded and x's diverge.
Each time x gets larger, 1 + 1/x gets smaller and (1 + 1/y)(1 + 1/z)
gets larger but only finite many times because each time x,y change
and there's just finite many ways they can change.
Thus the x,y,z's of solutions are all bounded
resulting in only finite many possible solutions.
answered Nov 16 at 9:10
William Elliot
6,7802518
add a comment |
up vote
0
down vote
up vote
0
down vote
x,y,z positive integers.
(1 + 1/x)(1 + 1/y)(1 + 1/z) = r.
Often there will be no solution. r = 7 for example.
Assume that x,y and z's of the solutions diverge.
That forces r = 1, contradicting the assumption of solutions.
Wlog assume the z's are bounded and the x,y's of solutions diverge.
For each solution, there's another solution with larger x,y.
Thus (1 + 1/x)(1 + 1/y) is smaller, 1 + 1/z is larger and z is
smaller. But z can be smaller only finite many times.
Wlog, assume y,z's of the solutions are bounded and x's diverge.
Each time x gets larger, 1 + 1/x gets smaller and (1 + 1/y)(1 + 1/z)
gets larger but only finite many times because each time x,y change
and there's just finite many ways they can change.
Thus the x,y,z's of solutions are all bounded
resulting in only finite many possible solutions.
answered Nov 16 at 9:10
William Elliot
6,7802518
x,y,z positive integers.
(1 + 1/x)(1 + 1/y)(1 + 1/z) = r.
Often there will be no solution. r = 7 for example.
Assume that x,y and z's of the solutions diverge.
That forces r = 1, contradicting the assumption of solutions.
Wlog assume the z's are bounded and the x,y's of solutions diverge.
For each solution, there's another solution with larger x,y.
Thus (1 + 1/x)(1 + 1/y) is smaller, 1 + 1/z is larger and z is
smaller. But z can be smaller only finite many times.
Wlog, assume y,z's of the solutions are bounded and x's diverge.
Each time x gets larger, 1 + 1/x gets smaller and (1 + 1/y)(1 + 1/z)
gets larger but only finite many times because each time x,y change
and there's just finite many ways they can change.
Thus the x,y,z's of solutions are all bounded
resulting in only finite many possible solutions.
answered Nov 16 at 9:10
William Elliot
6,7802518
answered Nov 16 at 9:10
William Elliot
6,7802518
answered Nov 16 at 9:10
William Elliot
6,7802518
answered Nov 16 at 9:10
William Elliot
6,7802518
6,7802518
add a comment |
add a comment |
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*For $ r=0 $?
– Raptor
Nov 11 at 11:54
1
@Raptor rational number greater than 1
– KGSH bteam Mine Team Beast O_
Nov 11 at 12:09
Show m(x + 1)(y + 1)(z + 1) = nxyz has finitely many x,y,z solutions for natural numbers m,n,x,y,z.
– William Elliot
Nov 11 at 22:16