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

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

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

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

• 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