# Number of solution of a rational sum

Let $$f(x)=\sum\limits_{i=1}^{2020}\frac{i^2}{x-i}$$. Then, what is the number of solutions of $$f(x)=0$$?

I am struck at this problem. Specifically, would not $$f(x)$$ become undefined when $$i=x$$. Since the numerator is always positive, therefore the sum becomes zero only when there are points when $$x>i$$ and $$x. Then, would not a point when $$x=i$$ appear? Any hints? Thanks beforehand.

## 1 Answer

Multiplying both sides by $$(x-1)\cdot\ldots\cdot(x-2020)$$ and denoting $$g(x)$$, what you get is:

$$g(x) = \sum_{i=0}^{2020} g_i(x)$$

where $$g_i(x)$$ is a polynomial of degree $$2019$$ such that $$g_i(n)=0$$ for all $$n\neq i$$ and $$g_i(i) = i!(2020-i)! \cdot (-1)^i i$$. Hence, you have that $$g(i)$$ and $$g(i+1)$$ alternate in sign. This says by continuity of $$g$$ that there is a root in every interval $$(i,i+1)$$ for $$i=1,\ldots,2019$$. Thus, you have $$2019$$ roots, and $$g(x)$$ is a polynomial of degree $$2019$$. So you have all of them. Moreover, these are all the roots of your $$f$$, since we only introduced a numerator that is zero only in integers and our roots lie in open intervals $$(i,i+1)$$.