Prove that all roots of $\sum_{r=1}^{70} \frac{1}{x-r} =\frac{5}{4} $ are real

Prove that all roots of $$\displaystyle \sum_{r=1}^{70} \dfrac{1}{x-r} =\dfrac{5}{4} $$

are real

I encountered this question in my weekly test. I tried setting $\displaystyle \sum_{r=1}^{70} \dfrac{1}{x-r} = \dfrac{P'(x)}{P(x)} $ ; where $\displaystyle P(x) = \prod_{r=1}^{70} (x-r)$ and tried to use some inequality, but to no avail.

Can we also generalize this?

Find the condition such that

$$\displaystyle \sum_{r=1}^{n} \dfrac{1}{x-r} = a $$

has all real roots.

$n\in \mathbb{Z^+} \ ; \ a\in \mathbb{R}$

Solution 1:

Alternatively: suppose that the function has a complex root; call it $x+iy$; $y\neq 0$.

Then $$\sum_{r=1}^{70}\frac{1}{x-r+iy}=\frac{5}{4}$$

multiply by the conjugate to get

$$\sum_{r=1}^{70}\frac{x-r-iy}{(x-r)^2+y^2}=\frac{5}{4}$$

which implies

$$\operatorname{Im}\left( \sum_{r=1}^{70}\frac{x-r-iy}{(x-r)^2+y^2} \right)=0$$ or $$\sum_{r=1}^{70}\frac{-y}{(x-r)^2+y^2}=0$$ which is a contradiction since each term has the same sign and is nonzero.

Solution 2:

Hint: Note that the derivative of the left-hand side, when it exists, is negative. So our function is decreasing in any interval $(k,k+1)$ where $1\le k\le 70-1$.

In this interval, our function is very large positive when $x$ is a tiny bit larger than $k$, and very large negative when $x$ is a tiny bit smaller than $k+1$. So by the Intermediate Value Theorem our equation has a root between $k$ and $k+1$.