Diophantine equation: $|\sin a|=|\sin b|^c$

Question

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.

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. — Jyrki Lahtonen, Jan 20 '19 at 19:03
See the tag wiki. Restricting the variables to range over integers only does not turn this into a diophantine equation. — Jyrki Lahtonen, Jan 20 '19 at 19:05

score 1 · Answer 1 · answered Jan 20 '19 at 12:22

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.

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. — Jyrki Lahtonen, Jan 20 '19 at 19:10
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. — Jyrki Lahtonen, Jan 20 '19 at 19:14

score 0 · Answer 2 · answered Jan 20 '19 at 11:40

0

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

answered Jan 20 '19 at 11:40

Peter Foreman

19,947

1

Why? Can you give a proof? – Szeto Jan 20 '19 at 12:15
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. – Peter Foreman Jan 20 '19 at 12:19

Diophantine equation: $|\sin a|=|\sin b|^c$

2 Answers2