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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.

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

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up vote 18 down vote accepted

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


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.

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(+1) Amazing answer! Thanks a lot. – Henry Mar 21 at 12:35
One might mention that this is essentially the proof of the Gauss-Lucas Theorem – Martin R Mar 21 at 14:41
+1, but -- why do you structure it as a proof by contradiction? (If you remove the unnecessary assumption that $y \neq 0$, isn't it a perfectly valid direct proof that the imaginary part of any solution is 0?) – ruakh Mar 25 at 1:14
Good point. I guess my mind jumps to contradiction. – Elliot G Mar 25 at 1:20

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 $k+1$.

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This yields 69 real roots of the 70th degree polynomial but then the 70th must also be real because the co-efficients are real so non-real roots occur in conjugate pairs. – user254665 Mar 21 at 6:53
@user254665: I had left that out to leave something to do. Another way of identifying the $70$-th is to use the same argument on $(70,\infty)$. Just to the right of $70$ the function is very large, and it decreases to $0$, passing $5/4$ along the way. – André Nicolas Mar 21 at 7:30
Ok. Good presentation. – user254665 Mar 21 at 7:51
(+1) Nice approach. Thanks. – Henry Mar 21 at 12:37

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