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Suppose that $x\in\mathbb{R}^+$ and $n\in \mathbb{N}$. If $\cosh(nx)$ and $\cosh((n+1)x)$ are rational, can we show that $\cosh(x)$ is rational too?

I guess the following equalities should be useful:

$$\begin{eqnarray} \cosh(x) &=& \cosh((n+1)x−xn)\\ &=& \cosh((n+1)n).\cosh(nx) − \sinh((n+1)x).\sinh(nx)\\ &=& \cosh((n+1)n).\cosh(nx) − \sqrt{\cosh^2((n+1)x) − 1}.\sqrt{\cosh^2(nx) − 1}. \end{eqnarray}$$

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    $\begingroup$ Is the corresponding proposition true for the ordinary trigonometric functions? $\endgroup$ – Lubin Apr 28 '16 at 17:50
  • $\begingroup$ @Lubin that I do not know. However, the below accepted answer does not work for ordinary trig functions like $\cos$ or $\sin$. $\endgroup$ – Luckyluck63 Apr 28 '16 at 23:22
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Suppose you have an $u>1$ such that both $u^n+u^{-n}$ and $u^{n+1} + u^{-n-1}$ are rational.

Then $u^n$ satisfies the equation with rational coefficients $(T - u^n)(T-u^{-n}) = T^2 - (u^n + u^{-n})T + 1 = 0$, so $u^n$ is algebraic and its only possible conjugate (over $\Bbb Q$) is $u^{-n}$.

Similarly, $u^{n+1}$ is algebraic and its only possible conjugate is $u^{-n-1}$.

Now, $u = u^{n+1}/u^n$, so $u$ is algebraic and its only possible conjugates are among $u^{-1},u^{2n+1},u^{-2n-1}$.

If $u^{2n+1}$ really is a conjugate of $u$, then so is $u^{(2n+1)^2}$. But since $u^{(2n+1)^2}$ is larger than any of $u,u^{-1},u^{2n+1},u^{-2n-1}$, this is impossible.

Therefore, $u$ has only one possible conjugate ($u^{-1}$), and so $u+u^{-1}$ is rational.

Taking $u = \exp x$, this shows that if $\cosh (nx)$ and $\cosh((n+1)x)$ are rational, $\cosh x$ is rational.


You can easily generalize this to show that if $n,m$ are positive integers and $\cosh (nx)$ and $\cosh((n+m)x)$ are rational, then $\cosh(mx)$ is rational


In fact, since we can view $\cosh(nx)$ as polynomials in $\cosh(x)$, we can interpret $\Bbb Q(\cosh(nx),\cosh((n+1)x)$ as a subfield of $\Bbb Q(\cosh x)$. Luroth's theorem then shows that this subfield is $\Bbb Q(\cosh x)$ itself, so there is a rational fraction $f_n$ in two variables with rational coefficients such that $\cosh(x) = f_n(\cosh(nx),\cosh((n+1)x)$.

This immediately implies that for almost all $x \in \Bbb C$, $\cosh(x) \in \Bbb Q(\cosh(nx),\cosh((n+1)x)$ (there is a finite number of exceptions where the denominator and numerator of $f_n$ vanish and $\cosh x$ may be in a bigger field).

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    $\begingroup$ I think your argument about the extensions $\Bbb Q(\cosh x)\supset\Bbb Q(\cosh(nx))$ is the most convincing, but I don’t hink it’s quite complete. It seems clear to me that the field-extension degree in this case is $n$, and the fact that $\Bbb Q(\cosh(nx),\cosh((n+1)x))$ is sandwiched between fields of degree $n$ and $n+1$ beneath $\Bbb Q(\cosh x)$ shows immediately that $\Bbb Q(\cosh(nx),\cosh((n+1)x))=\Bbb Q(\cosh x)$. This argument works as well for the ordinary trigonometric functions, too. $\endgroup$ – Lubin Apr 29 '16 at 5:12

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