Consider the sum $S=\sum\frac{1}{x^2}$ which is over all the positive real solutions of the equation $\frac{\tan{x}}{x}=n$ Consider the sum $S(n)=\sum\dfrac{1}{x^2}$
where summation is performed over all the positive real solutions of the equation $\dfrac{\tan{x}}{x}=n$.
If it's given that $S(n)=1$, $n\in\mathbb{Q}$, find $n$.
My attempt:
We're interested in the roots of
$$\frac{\tan x}{x}=k$$
Then, performing Taylor expansion of
$$\sin x = kx\cos x$$
$$x- \frac{(x^6)}{6}+ ......  = kx(1- \frac{x^2}{2}+........$$
I am stuck here. Any hints will be appreciated.
Thanks
 A: The idea of the solution is due to (now deleted) answer of runway44. The solution is based on the comparison of Taylor series and Hadamard factorization of the function
$$
f_n(z)=\frac{\sin z}z-n\cos z,\tag1
$$
which are:
$$
f_n(z)=1-\frac16z^2+\dots-n\left(1-\frac12z^2+\dots\right)
=(1-n)-\left(\frac16-\frac n2\right)z^2+\dots
$$
and
$$
f_n(z)=f_n(0)\prod_{\rho}^{f_n(\rho)=0}\left(1-\frac z{\rho}\right)e^{z/\rho}
=f_n(0)\prod_{\rho_*}^{f_n(\rho_*)=0}\left(1-\frac {z^2}{\rho_*^2}\right)
=(1-n)\left(1-z^2\sum_{\rho_*}^{f_n(\rho_*)=0}\frac1{\rho_*^2}+\dots\right),
$$
respectively, where we used the fact that all roots $\rho$ (except for $0$ which requires $n=1$) are paired $f_n(\rho)=0\iff f_n(-\rho)=0$. Due to this fact we can effectively use one root $\rho_*$ from each pair. For convenience we may assume $0\le\operatorname{Arg}(\rho_*)<\pi$.
Comparing the coefficients at $z^2$ in the above expressions one obtains the equality:
$$%\sideset{}'
(1-n)\sum_{\rho_*}^{f_n(\rho_*)=0}\frac1{\rho_*^2}=\frac16-\frac n2.\tag2
$$
A fine detail here is the domain of summation. Whereas the sum in question is over the real roots, the sum in (2) is over all complex roots of the function $f_n(z)$. The two sets are not necessarily the same, as we will immediately see. Indeed:
$$\begin{align}
f_n(z)=0&\implies |\sin z-nz\cos z|^2=0\\
&\stackrel{z=x+iy}\implies (\sin x-nx\cos x)^2+(\sinh y-ny\cosh y)^2
+n^2x^2y^2\left(\frac{\sinh^2y}{y^2}-\frac{\sin^2x}{x^2}\right)=0.
\end{align}
$$
Obviously the last equality can hold only if $x=0$ or $y=0$:
$$\begin{align}
x=0:&\quad \tanh y=n y;\tag{3a}\\
y=0:&\quad \tan x=n x.\tag{3b}\\
\end{align}
$$
Whereas the equation (3b) delivers all non-zero real roots of the function $f_n(z)$ the equation (3a) results in an additional pair of imaginary roots provided that $0<n<1$ (otherwise (3a) has no non-zero real solutions).
Let the imaginary root of $f_n(z)$ be (if it exists) $iy_n$, where $y_n$ is the non-zero real solution of (3a). Then (2) will read:
$$
-\frac{\mathbb1_{0<n<1}}{y_n^2}+\sum_{\rho>0}^{f_n(\rho)=0}\frac1{\rho^2}=\frac{\frac16-\frac n2}{1-n}.\tag4
$$
Coming back to the original problem this means that the assumption $n\not\in(0,1)$ implies $n=\frac53$. However some additional work is required to show that there is no rational $n$ in the range $(0,1)$ that would satisfy the problem.
In fact, the condition $n\in\mathbb{Q}$ appears to be excessive, since quite obviously $S(n)\le S(0)=\frac16$ for $0\le n\le1$. Generally, it can be shown that the problem has a unique solution $n$ for any $S(n)\in[\frac1{10},\frac12)\cup(\frac12,+\infty)$, whereas there is no solution for any other value.
A reasonable variation of the problem would be asking the same question for $S(n)=\frac18$ (or any other rational value between $\frac1{10}$ and $\frac16$). The correct answer would be that there exists no rational $n$ satisfying the problem, which can be shown by applying Lindemann–Weierstrass theorem.
