0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 '19 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 '19 at 19:05
1
$\begingroup$

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 $c\in\mathbb{Z}\setminus\{0\}$ 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 $c\neq0,1$? Clearly $a\neq b$ 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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ The values of sines are obviously not all algebraically independent. We have, after all, formulas like $$\sin 3x=3\sin x-4\sin^3x,$$ creating algebraic dependencies between $\sin1$ and $\sin 3$, $\sin 2$ and $\sin 6$ etc. $\endgroup$ – Jyrki Lahtonen Jan 20 '19 at 19:10
  • $\begingroup$ Also $$\sin^22x=4\sin^2x\cos^2x=4\sin^2x(1-\sin^2x).$$ 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 '19 at 19:14
0
$\begingroup$

The only other solutions would be $a=b=0$ and $c \in \mathbb{Z}$.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Why? Can you give a proof? $\endgroup$ – Szeto Jan 20 '19 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 '19 at 12:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.