How many real solutions does the equation $f(x) =0$ have, where $f(x) =\sum_{i=1}^{2020} \frac {i^2}{\left( x-i\right)} $ How many real solutions does the equation $f(x) =0$ have, where
$f(x) =\sum_{i=1}^{2020} \frac {i^2}{\left( x-i\right)} $
$\bf{My Try} :$  This is a equation of degree 2020. Now let $\alpha$ be a complex root. Then we have
$\sum_{i=1}^{2020} \frac {i^2}{\left( \alpha -i\right)}=0 $
And $\sum_{i=1}^{2020} \frac {i^2}{\left( \bar {\alpha} -i\right)}=0 $
Now subtracting the 2nd from the 1st  equation we get
$\sum_{i=1}^{2020} i^2\frac {\bar \alpha - \alpha }{\left( \alpha -i\right)\left(\bar \alpha - i\right) }=0 $
Hence $\bar \alpha - \alpha =0 \implies \alpha$ is real
Hence the equation has 2020 real solutions.
$\bf\text{  Is my argument correct?  
  Is there any alternative way to solve this problem?}$
Thanks in advance.
 A: First off, the equation is of degree $2019$, not $2020$. But besides this, I think you could use more details as to why
$$\sum_{i=1}^{2020} i^2\frac {\bar \alpha - \alpha }{\left( \alpha -i\right)\left(\bar \alpha - i\right) }=0$$
implies
$$\bar{\alpha}-\alpha=0$$
Could it not be the case that
$$\sum_{i=1}^{2020} \frac {i^2}{\left( \alpha -i\right)\left(\bar \alpha - i\right) }=0$$
It turns out this can not be the case, but at least to my eyes it is not obvious. Let $\alpha=x+iy$. Then
$$(\bar{\alpha}-n)(\alpha-n)=(x-iy-n)(x+iy-n)$$
$$=n^2-2nx+x^2+y^2=(n-x)^2+y^2\geq y^2>0$$
Note that I have replaced $i$ in the summation with $n$ (and used $i=\sqrt{-1}$) and that $y\neq 0$ since $\alpha\not\in\mathbb{R}$. This then implies
$$\sum_{i=1}^{2020} \frac {i^2}{\left( \alpha -i\right)\left(\bar \alpha - i\right) }>\sum_{i=1}^{2020}0=0$$
So your reasoning is correct, but you could definitely use some more details.

As an aside, this is actually a great example of a good use of contradiction (at least this method of proving it). You start by assuming that such a non-real $\alpha$ exists, which then implies that $\alpha$ is real. I just found that interesting because usually contradiction ends up with $0=1$ or something of that nature. Here, we contradict our original assumption which some would argue is the only valid use of contradiction (math philosophers prepare your attacks).
