Mixing
Diophantine equation: $|sin a|=|sin b|^c$
$begingroup$
Does there exist integer solutions to
$$|sin a|=|sin b|^c$$
other than $a=b$, $c=1$?
Currently I have no progress. To merely satisfy the requirements of MSE, I can only say that I invented this problem when I try to create Diophantine equations that involve special functions.
I apologize for that.
Thanks for any help in advance.
diophantine-equations
asked Jan 20 at 11:35
SzetoSzeto
6,5912926
$endgroup$
add a comment |
$begingroup$
Does there exist integer solutions to
$$|sin a|=|sin b|^c$$
other than $a=b$, $c=1$?
Currently I have no progress. To merely satisfy the requirements of MSE, I can only say that I invented this problem when I try to create Diophantine equations that involve special functions.
I apologize for that.
Thanks for any help in advance.
diophantine-equations
asked Jan 20 at 11:35
SzetoSzeto
6,5912926
$endgroup$
$begingroup$
I think that the act of introducing the sine immediately took you away from the realm of diophantine equations. Only powers, sums and products qualify I think.
$endgroup$
– Jyrki Lahtonen
Jan 20 at 19:03
$begingroup$
See the tag wiki. Restricting the variables to range over integers only does not turn this into a diophantine equation.
$endgroup$
– Jyrki Lahtonen
Jan 20 at 19:05
add a comment |
$begingroup$
Does there exist integer solutions to
$$|sin a|=|sin b|^c$$
other than $a=b$, $c=1$?
Currently I have no progress. To merely satisfy the requirements of MSE, I can only say that I invented this problem when I try to create Diophantine equations that involve special functions.
I apologize for that.
Thanks for any help in advance.
diophantine-equations
asked Jan 20 at 11:35
SzetoSzeto
6,5912926
$endgroup$
Does there exist integer solutions to
$$|sin a|=|sin b|^c$$
other than $a=b$, $c=1$?
Currently I have no progress. To merely satisfy the requirements of MSE, I can only say that I invented this problem when I try to create Diophantine equations that involve special functions.
I apologize for that.
Thanks for any help in advance.
diophantine-equations
diophantine-equations
asked Jan 20 at 11:35
SzetoSzeto
6,5912926
asked Jan 20 at 11:35
SzetoSzeto
6,5912926
asked Jan 20 at 11:35
SzetoSzeto
6,5912926
asked Jan 20 at 11:35
SzetoSzeto
6,5912926
asked Jan 20 at 11:35
SzetoSzeto
6,5912926
6,5912926
$begingroup$
I think that the act of introducing the sine immediately took you away from the realm of diophantine equations. Only powers, sums and products qualify I think.
$endgroup$
– Jyrki Lahtonen
Jan 20 at 19:03
$begingroup$
See the tag wiki. Restricting the variables to range over integers only does not turn this into a diophantine equation.
$endgroup$
– Jyrki Lahtonen
Jan 20 at 19:05
add a comment |
$begingroup$
I think that the act of introducing the sine immediately took you away from the realm of diophantine equations. Only powers, sums and products qualify I think.
$endgroup$
– Jyrki Lahtonen
Jan 20 at 19:03
$begingroup$
See the tag wiki. Restricting the variables to range over integers only does not turn this into a diophantine equation.
$endgroup$
– Jyrki Lahtonen
Jan 20 at 19:05
$begingroup$
I think that the act of introducing the sine immediately took you away from the realm of diophantine equations. Only powers, sums and products qualify I think.
$endgroup$
– Jyrki Lahtonen
Jan 20 at 19:03
$begingroup$
I think that the act of introducing the sine immediately took you away from the realm of diophantine equations. Only powers, sums and products qualify I think.
$endgroup$
– Jyrki Lahtonen
Jan 20 at 19:03
$begingroup$
See the tag wiki. Restricting the variables to range over integers only does not turn this into a diophantine equation.
$endgroup$
– Jyrki Lahtonen
Jan 20 at 19:05
$begingroup$
See the tag wiki. Restricting the variables to range over integers only does not turn this into a diophantine equation.
$endgroup$
– Jyrki Lahtonen
Jan 20 at 19:05
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
For the diophantine equation
$$
|sin(a)|=|sin(b)|^c
$$
there are some obvious classes of solutions:
1) Take $a=b$ and $c=1$, as in the OP.
2) Take $a=-b$ and $c=1$.
3) Take $a=b=0$ and $cinmathbb{Z}setminus{0}$ as in another answer.
What else can we say? We know that $sin(x)$ is transcendental at non-zero integer values, so $c=0$ can have no solutions. I will leave $0^0$ undefined, but if you do define it you might get an extra solution $(0=0^0)$.
Can there be any other solutions for $cneq0,1$? Clearly $aneq b$ is required, but beyond this (if I recall my transcendental number theory correctly) I think you have an open problem. We do not know that the values of $sin(x)$ at different integers are algebraically independent, so there might be a solution or there might not be.
answered Jan 20 at 12:22
GoodQuestionBroGoodQuestionBro
111
$endgroup$
$begingroup$
The values of sines are obviously not all algebraically independent. We have, after all, formulas like $$sin 3x=3sin x-4sin^3x,$$ creating algebraic dependencies between $sin1$ and $sin 3$, $sin 2$ and $sin 6$ etc.
$endgroup$
– Jyrki Lahtonen
Jan 20 at 19:10
$begingroup$
Also $$sin^22x=4sin^2xcos^2x=4sin^2x(1-sin^2x).$$ Looks like any two numbers $sin a$ and $sin b$ with $a,b$ non-zero integers are algebraically dependent. Using complex exponential makes this even more obvious.
$endgroup$
– Jyrki Lahtonen
Jan 20 at 19:14
add a comment |
$begingroup$
The only other solutions would be $a=b=0$ and $c in mathbb{Z}$.
answered Jan 20 at 11:40
Peter ForemanPeter Foreman
2,7421214
$endgroup$
$begingroup$
Why? Can you give a proof?
$endgroup$
– Szeto
Jan 20 at 12:15
$begingroup$
Sine applied to an integer will always result in a non rational number. I do not know how to prove it, but I am quite sure that no value of $sin{(a)}$ will be a surd such that $sin{(b)}$ will be the $c$th root of $sin{(a)}$. So the only possible solutions are as you stated or where both sides of the equation are 0.
$endgroup$
– Peter Foreman
Jan 20 at 12:19
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For the diophantine equation
$$
|sin(a)|=|sin(b)|^c
$$
there are some obvious classes of solutions:
1) Take $a=b$ and $c=1$, as in the OP.
2) Take $a=-b$ and $c=1$.
3) Take $a=b=0$ and $cinmathbb{Z}setminus{0}$ as in another answer.
What else can we say? We know that $sin(x)$ is transcendental at non-zero integer values, so $c=0$ can have no solutions. I will leave $0^0$ undefined, but if you do define it you might get an extra solution $(0=0^0)$.
Can there be any other solutions for $cneq0,1$? Clearly $aneq b$ is required, but beyond this (if I recall my transcendental number theory correctly) I think you have an open problem. We do not know that the values of $sin(x)$ at different integers are algebraically independent, so there might be a solution or there might not be.
answered Jan 20 at 12:22
GoodQuestionBroGoodQuestionBro
111
$endgroup$
$begingroup$
The values of sines are obviously not all algebraically independent. We have, after all, formulas like $$sin 3x=3sin x-4sin^3x,$$ creating algebraic dependencies between $sin1$ and $sin 3$, $sin 2$ and $sin 6$ etc.
$endgroup$
– Jyrki Lahtonen
Jan 20 at 19:10
$begingroup$
Also $$sin^22x=4sin^2xcos^2x=4sin^2x(1-sin^2x).$$ Looks like any two numbers $sin a$ and $sin b$ with $a,b$ non-zero integers are algebraically dependent. Using complex exponential makes this even more obvious.
$endgroup$
– Jyrki Lahtonen
Jan 20 at 19:14
add a comment |
$begingroup$
For the diophantine equation
$$
|sin(a)|=|sin(b)|^c
$$
there are some obvious classes of solutions:
1) Take $a=b$ and $c=1$, as in the OP.
2) Take $a=-b$ and $c=1$.
3) Take $a=b=0$ and $cinmathbb{Z}setminus{0}$ as in another answer.
What else can we say? We know that $sin(x)$ is transcendental at non-zero integer values, so $c=0$ can have no solutions. I will leave $0^0$ undefined, but if you do define it you might get an extra solution $(0=0^0)$.
Can there be any other solutions for $cneq0,1$? Clearly $aneq b$ is required, but beyond this (if I recall my transcendental number theory correctly) I think you have an open problem. We do not know that the values of $sin(x)$ at different integers are algebraically independent, so there might be a solution or there might not be.
answered Jan 20 at 12:22
GoodQuestionBroGoodQuestionBro
111
$endgroup$
$begingroup$
The values of sines are obviously not all algebraically independent. We have, after all, formulas like $$sin 3x=3sin x-4sin^3x,$$ creating algebraic dependencies between $sin1$ and $sin 3$, $sin 2$ and $sin 6$ etc.
$endgroup$
– Jyrki Lahtonen
Jan 20 at 19:10
$begingroup$
Also $$sin^22x=4sin^2xcos^2x=4sin^2x(1-sin^2x).$$ Looks like any two numbers $sin a$ and $sin b$ with $a,b$ non-zero integers are algebraically dependent. Using complex exponential makes this even more obvious.
$endgroup$
– Jyrki Lahtonen
Jan 20 at 19:14
add a comment |
$begingroup$
For the diophantine equation
$$
|sin(a)|=|sin(b)|^c
$$
there are some obvious classes of solutions:
1) Take $a=b$ and $c=1$, as in the OP.
2) Take $a=-b$ and $c=1$.
3) Take $a=b=0$ and $cinmathbb{Z}setminus{0}$ as in another answer.
What else can we say? We know that $sin(x)$ is transcendental at non-zero integer values, so $c=0$ can have no solutions. I will leave $0^0$ undefined, but if you do define it you might get an extra solution $(0=0^0)$.
Can there be any other solutions for $cneq0,1$? Clearly $aneq b$ is required, but beyond this (if I recall my transcendental number theory correctly) I think you have an open problem. We do not know that the values of $sin(x)$ at different integers are algebraically independent, so there might be a solution or there might not be.
answered Jan 20 at 12:22
GoodQuestionBroGoodQuestionBro
111
$endgroup$
For the diophantine equation
$$
|sin(a)|=|sin(b)|^c
$$
there are some obvious classes of solutions:
1) Take $a=b$ and $c=1$, as in the OP.
2) Take $a=-b$ and $c=1$.
3) Take $a=b=0$ and $cinmathbb{Z}setminus{0}$ as in another answer.
What else can we say? We know that $sin(x)$ is transcendental at non-zero integer values, so $c=0$ can have no solutions. I will leave $0^0$ undefined, but if you do define it you might get an extra solution $(0=0^0)$.
Can there be any other solutions for $cneq0,1$? Clearly $aneq b$ is required, but beyond this (if I recall my transcendental number theory correctly) I think you have an open problem. We do not know that the values of $sin(x)$ at different integers are algebraically independent, so there might be a solution or there might not be.
answered Jan 20 at 12:22
GoodQuestionBroGoodQuestionBro
111
answered Jan 20 at 12:22
GoodQuestionBroGoodQuestionBro
111
answered Jan 20 at 12:22
GoodQuestionBroGoodQuestionBro
111
answered Jan 20 at 12:22
GoodQuestionBroGoodQuestionBro
111
111
$begingroup$
The values of sines are obviously not all algebraically independent. We have, after all, formulas like $$sin 3x=3sin x-4sin^3x,$$ creating algebraic dependencies between $sin1$ and $sin 3$, $sin 2$ and $sin 6$ etc.
$endgroup$
– Jyrki Lahtonen
Jan 20 at 19:10
$begingroup$
Also $$sin^22x=4sin^2xcos^2x=4sin^2x(1-sin^2x).$$ Looks like any two numbers $sin a$ and $sin b$ with $a,b$ non-zero integers are algebraically dependent. Using complex exponential makes this even more obvious.
$endgroup$
– Jyrki Lahtonen
Jan 20 at 19:14
add a comment |
$begingroup$
The values of sines are obviously not all algebraically independent. We have, after all, formulas like $$sin 3x=3sin x-4sin^3x,$$ creating algebraic dependencies between $sin1$ and $sin 3$, $sin 2$ and $sin 6$ etc.
$endgroup$
– Jyrki Lahtonen
Jan 20 at 19:10
$begingroup$
Also $$sin^22x=4sin^2xcos^2x=4sin^2x(1-sin^2x).$$ Looks like any two numbers $sin a$ and $sin b$ with $a,b$ non-zero integers are algebraically dependent. Using complex exponential makes this even more obvious.
$endgroup$
– Jyrki Lahtonen
Jan 20 at 19:14
$begingroup$
The values of sines are obviously not all algebraically independent. We have, after all, formulas like $$sin 3x=3sin x-4sin^3x,$$ creating algebraic dependencies between $sin1$ and $sin 3$, $sin 2$ and $sin 6$ etc.
$endgroup$
– Jyrki Lahtonen
Jan 20 at 19:10
$begingroup$
The values of sines are obviously not all algebraically independent. We have, after all, formulas like $$sin 3x=3sin x-4sin^3x,$$ creating algebraic dependencies between $sin1$ and $sin 3$, $sin 2$ and $sin 6$ etc.
$endgroup$
– Jyrki Lahtonen
Jan 20 at 19:10
$begingroup$
Also $$sin^22x=4sin^2xcos^2x=4sin^2x(1-sin^2x).$$ Looks like any two numbers $sin a$ and $sin b$ with $a,b$ non-zero integers are algebraically dependent. Using complex exponential makes this even more obvious.
$endgroup$
– Jyrki Lahtonen
Jan 20 at 19:14
$begingroup$
Also $$sin^22x=4sin^2xcos^2x=4sin^2x(1-sin^2x).$$ Looks like any two numbers $sin a$ and $sin b$ with $a,b$ non-zero integers are algebraically dependent. Using complex exponential makes this even more obvious.
$endgroup$
– Jyrki Lahtonen
Jan 20 at 19:14
add a comment |
$begingroup$
The only other solutions would be $a=b=0$ and $c in mathbb{Z}$.
answered Jan 20 at 11:40
Peter ForemanPeter Foreman
2,7421214
$endgroup$
$begingroup$
Why? Can you give a proof?
$endgroup$
– Szeto
Jan 20 at 12:15
$begingroup$
Sine applied to an integer will always result in a non rational number. I do not know how to prove it, but I am quite sure that no value of $sin{(a)}$ will be a surd such that $sin{(b)}$ will be the $c$th root of $sin{(a)}$. So the only possible solutions are as you stated or where both sides of the equation are 0.
$endgroup$
– Peter Foreman
Jan 20 at 12:19
add a comment |
$begingroup$
The only other solutions would be $a=b=0$ and $c in mathbb{Z}$.
answered Jan 20 at 11:40
Peter ForemanPeter Foreman
2,7421214
$endgroup$
$begingroup$
Why? Can you give a proof?
$endgroup$
– Szeto
Jan 20 at 12:15
$begingroup$
Sine applied to an integer will always result in a non rational number. I do not know how to prove it, but I am quite sure that no value of $sin{(a)}$ will be a surd such that $sin{(b)}$ will be the $c$th root of $sin{(a)}$. So the only possible solutions are as you stated or where both sides of the equation are 0.
$endgroup$
– Peter Foreman
Jan 20 at 12:19
add a comment |
$begingroup$
The only other solutions would be $a=b=0$ and $c in mathbb{Z}$.
answered Jan 20 at 11:40
Peter ForemanPeter Foreman
2,7421214
$endgroup$
The only other solutions would be $a=b=0$ and $c in mathbb{Z}$.
answered Jan 20 at 11:40
Peter ForemanPeter Foreman
2,7421214
answered Jan 20 at 11:40
Peter ForemanPeter Foreman
2,7421214
answered Jan 20 at 11:40
Peter ForemanPeter Foreman
2,7421214
answered Jan 20 at 11:40
Peter ForemanPeter Foreman
2,7421214
2,7421214
$begingroup$
Why? Can you give a proof?
$endgroup$
– Szeto
Jan 20 at 12:15
$begingroup$
Sine applied to an integer will always result in a non rational number. I do not know how to prove it, but I am quite sure that no value of $sin{(a)}$ will be a surd such that $sin{(b)}$ will be the $c$th root of $sin{(a)}$. So the only possible solutions are as you stated or where both sides of the equation are 0.
$endgroup$
– Peter Foreman
Jan 20 at 12:19
add a comment |
$begingroup$
Why? Can you give a proof?
$endgroup$
– Szeto
Jan 20 at 12:15
$begingroup$
Sine applied to an integer will always result in a non rational number. I do not know how to prove it, but I am quite sure that no value of $sin{(a)}$ will be a surd such that $sin{(b)}$ will be the $c$th root of $sin{(a)}$. So the only possible solutions are as you stated or where both sides of the equation are 0.
$endgroup$
– Peter Foreman
Jan 20 at 12:19
$begingroup$
Why? Can you give a proof?
$endgroup$
– Szeto
Jan 20 at 12:15
$begingroup$
Why? Can you give a proof?
$endgroup$
– Szeto
Jan 20 at 12:15
$begingroup$
Sine applied to an integer will always result in a non rational number. I do not know how to prove it, but I am quite sure that no value of $sin{(a)}$ will be a surd such that $sin{(b)}$ will be the $c$th root of $sin{(a)}$. So the only possible solutions are as you stated or where both sides of the equation are 0.
$endgroup$
– Peter Foreman
Jan 20 at 12:19
$begingroup$
Sine applied to an integer will always result in a non rational number. I do not know how to prove it, but I am quite sure that no value of $sin{(a)}$ will be a surd such that $sin{(b)}$ will be the $c$th root of $sin{(a)}$. So the only possible solutions are as you stated or where both sides of the equation are 0.
$endgroup$
– Peter Foreman
Jan 20 at 12:19
add a comment |
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$begingroup$
I think that the act of introducing the sine immediately took you away from the realm of diophantine equations. Only powers, sums and products qualify I think.
$endgroup$
– Jyrki Lahtonen
Jan 20 at 19:03
$begingroup$
See the tag wiki. Restricting the variables to range over integers only does not turn this into a diophantine equation.
$endgroup$
– Jyrki Lahtonen
Jan 20 at 19:05