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Are there natural number satisfying the following equations?
Clash Royale CLAN TAG#URR8PPP
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The motivation comes from the following question on MathOverflow:
Is the exponential version of Catalan-Dickson conjecture true?
Question 1. Is there a natural number $n$ satisfying the equation $n(n+1)=2^[log^n_2]+2$ where $[.]$ indicates the floor function. I am also interested in the minimum $n$ satisfying this equation if there is any.
As the answer to the above question is negative, let's consider the following more generalized form:
Question 2. Are there natural number $n$ and prime number $p$ satisfying the equation $fracn(n+1)2=fracpp-1times p^[log_p^n]+fracp-2p-1times p^n$ where $[.]$ indicates the floor function. (Note that for $p=2$ we get the previous equation.)
number-theory discrete-mathematics logarithms natural-numbers
asked Jul 21 at 9:27
Morteza Azad
1
add a comment |Â
up vote
0
down vote
favorite
The motivation comes from the following question on MathOverflow:
Is the exponential version of Catalan-Dickson conjecture true?
Question 1. Is there a natural number $n$ satisfying the equation $n(n+1)=2^[log^n_2]+2$ where $[.]$ indicates the floor function. I am also interested in the minimum $n$ satisfying this equation if there is any.
As the answer to the above question is negative, let's consider the following more generalized form:
Question 2. Are there natural number $n$ and prime number $p$ satisfying the equation $fracn(n+1)2=fracpp-1times p^[log_p^n]+fracp-2p-1times p^n$ where $[.]$ indicates the floor function. (Note that for $p=2$ we get the previous equation.)
number-theory discrete-mathematics logarithms natural-numbers
asked Jul 21 at 9:27
Morteza Azad
1
1
Well for $n>1$ the left hand side would not be a power of $2$, so no.
– Gal Porat
Jul 21 at 9:33
1
Anyway, what is $log_2^n$?
– Hagen von Eitzen
Jul 21 at 9:35
For the more general question, $fracp-2p-1cdot p^n$ quickly outgrows $fracn(n+1)2$...
– Wojowu
Jul 21 at 10:43
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
The motivation comes from the following question on MathOverflow:
Is the exponential version of Catalan-Dickson conjecture true?
Question 1. Is there a natural number $n$ satisfying the equation $n(n+1)=2^[log^n_2]+2$ where $[.]$ indicates the floor function. I am also interested in the minimum $n$ satisfying this equation if there is any.
As the answer to the above question is negative, let's consider the following more generalized form:
Question 2. Are there natural number $n$ and prime number $p$ satisfying the equation $fracn(n+1)2=fracpp-1times p^[log_p^n]+fracp-2p-1times p^n$ where $[.]$ indicates the floor function. (Note that for $p=2$ we get the previous equation.)
number-theory discrete-mathematics logarithms natural-numbers
asked Jul 21 at 9:27
Morteza Azad
1
The motivation comes from the following question on MathOverflow:
Is the exponential version of Catalan-Dickson conjecture true?
Question 1. Is there a natural number $n$ satisfying the equation $n(n+1)=2^[log^n_2]+2$ where $[.]$ indicates the floor function. I am also interested in the minimum $n$ satisfying this equation if there is any.
As the answer to the above question is negative, let's consider the following more generalized form:
Question 2. Are there natural number $n$ and prime number $p$ satisfying the equation $fracn(n+1)2=fracpp-1times p^[log_p^n]+fracp-2p-1times p^n$ where $[.]$ indicates the floor function. (Note that for $p=2$ we get the previous equation.)
number-theory discrete-mathematics logarithms natural-numbers
asked Jul 21 at 9:27
Morteza Azad
1
edited Jul 21 at 10:36
asked Jul 21 at 9:27
Morteza Azad
1
asked Jul 21 at 9:27
Morteza Azad
1
asked Jul 21 at 9:27
Morteza Azad
1
1
1
Well for $n>1$ the left hand side would not be a power of $2$, so no.
– Gal Porat
Jul 21 at 9:33
1
Anyway, what is $log_2^n$?
– Hagen von Eitzen
Jul 21 at 9:35
For the more general question, $fracp-2p-1cdot p^n$ quickly outgrows $fracn(n+1)2$...
– Wojowu
Jul 21 at 10:43
add a comment |Â
1
Well for $n>1$ the left hand side would not be a power of $2$, so no.
– Gal Porat
Jul 21 at 9:33
1
Anyway, what is $log_2^n$?
– Hagen von Eitzen
Jul 21 at 9:35
For the more general question, $fracp-2p-1cdot p^n$ quickly outgrows $fracn(n+1)2$...
– Wojowu
Jul 21 at 10:43
1
1
Well for $n>1$ the left hand side would not be a power of $2$, so no.
– Gal Porat
Jul 21 at 9:33
Well for $n>1$ the left hand side would not be a power of $2$, so no.
– Gal Porat
Jul 21 at 9:33
1
1
Anyway, what is $log_2^n$?
– Hagen von Eitzen
Jul 21 at 9:35
Anyway, what is $log_2^n$?
– Hagen von Eitzen
Jul 21 at 9:35
For the more general question, $fracp-2p-1cdot p^n$ quickly outgrows $fracn(n+1)2$...
– Wojowu
Jul 21 at 10:43
For the more general question, $fracp-2p-1cdot p^n$ quickly outgrows $fracn(n+1)2$...
– Wojowu
Jul 21 at 10:43
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
The RHS of $$n(n+1)=2^[log^n_2]+2$$ is a power of $2$ while the LHS is not a power of $2$ unless $n=1$ and $n=1$ is not a solution.
Thus I would say there are no solutions for this equation.
answered Jul 21 at 10:13
Mohammad Riazi-Kermani
27.5k41852
And for the second question the left side is an integer while the right side is probably not. It is difficult to get rid of the $p-1$ in the denominator.
– Ross Millikan
Jul 21 at 20:51
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
The RHS of $$n(n+1)=2^[log^n_2]+2$$ is a power of $2$ while the LHS is not a power of $2$ unless $n=1$ and $n=1$ is not a solution.
Thus I would say there are no solutions for this equation.
answered Jul 21 at 10:13
Mohammad Riazi-Kermani
27.5k41852
And for the second question the left side is an integer while the right side is probably not. It is difficult to get rid of the $p-1$ in the denominator.
– Ross Millikan
Jul 21 at 20:51
add a comment |Â
up vote
3
down vote
accepted
The RHS of $$n(n+1)=2^[log^n_2]+2$$ is a power of $2$ while the LHS is not a power of $2$ unless $n=1$ and $n=1$ is not a solution.
Thus I would say there are no solutions for this equation.
answered Jul 21 at 10:13
Mohammad Riazi-Kermani
27.5k41852
And for the second question the left side is an integer while the right side is probably not. It is difficult to get rid of the $p-1$ in the denominator.
– Ross Millikan
Jul 21 at 20:51
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
The RHS of $$n(n+1)=2^[log^n_2]+2$$ is a power of $2$ while the LHS is not a power of $2$ unless $n=1$ and $n=1$ is not a solution.
Thus I would say there are no solutions for this equation.
answered Jul 21 at 10:13
Mohammad Riazi-Kermani
27.5k41852
The RHS of $$n(n+1)=2^[log^n_2]+2$$ is a power of $2$ while the LHS is not a power of $2$ unless $n=1$ and $n=1$ is not a solution.
Thus I would say there are no solutions for this equation.
answered Jul 21 at 10:13
Mohammad Riazi-Kermani
27.5k41852
edited Jul 21 at 19:51
answered Jul 21 at 10:13
Mohammad Riazi-Kermani
27.5k41852
answered Jul 21 at 10:13
Mohammad Riazi-Kermani
27.5k41852
answered Jul 21 at 10:13
Mohammad Riazi-Kermani
27.5k41852
27.5k41852
And for the second question the left side is an integer while the right side is probably not. It is difficult to get rid of the $p-1$ in the denominator.
– Ross Millikan
Jul 21 at 20:51
add a comment |Â
And for the second question the left side is an integer while the right side is probably not. It is difficult to get rid of the $p-1$ in the denominator.
– Ross Millikan
Jul 21 at 20:51
And for the second question the left side is an integer while the right side is probably not. It is difficult to get rid of the $p-1$ in the denominator.
– Ross Millikan
Jul 21 at 20:51
And for the second question the left side is an integer while the right side is probably not. It is difficult to get rid of the $p-1$ in the denominator.
– Ross Millikan
Jul 21 at 20:51
add a comment |Â
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1
Well for $n>1$ the left hand side would not be a power of $2$, so no.
– Gal Porat
Jul 21 at 9:33
1
Anyway, what is $log_2^n$?
– Hagen von Eitzen
Jul 21 at 9:35
For the more general question, $fracp-2p-1cdot p^n$ quickly outgrows $fracn(n+1)2$...
– Wojowu
Jul 21 at 10:43