How many real solutions does the equation $f(x)=0$ have, where $f(x)$ is defined as follows?

$$f(x) = \sum\limits_{i=1}^{2020}\frac{i^2}{x-i}.$$

Now, by clearing the denominator in $f(x)$, we have $$f(x) = \sum\limits_{i=1}^{2020} f_i(x).$$ This shows that $f(x)$ is a polynomial of degree $2019$. From the above discussions, we observe that, for $1 \le i \le 2020$, $$f(i) = f_i(i).$$

This shows that, the function $f$ changes its sign when we go from $1$ to $2020$ by one integer to next integer. Hence, the function $f(x)$ has atleast one real zero in each of the interval $(i,i+1)$ ($1\le i \le2019$) and in total it has atleast 2019 real zeros. But $f(x)$ itself is a polynomial of degree $2019$ and hence this should be the full set of real roots of $f(x)$.

$$f(x) = \sum\limits_{i=1}^{2020}\frac{i^2}{x-i}.$$

**Solution**: For each $1 \le i \le 2020$ consider the following polynomial of degree 2019: $$f_i (x):= i^2 \prod\limits_{\substack{j=1 \\ j \ne i}}^{2020} (x-j).$$ Let $1 \le k \le 2020$, then $f_i(k) = 0$ if $i \ne k$ and $f_i(i) = i^2 \prod\limits_{\substack{j=1 \\ j \ne i}}^{2020}(i-j) = i^2 \prod\limits_{j=1}^{i-1} (i-j) \prod\limits_{j=i+1}^{2020} (i-j)$. Hence, the sign of $f_i(i) = (-1)^{2020-i}$. This shows that the sign of the numbers $f_1(1),f_2(2),\dots,f_{2020}(2020)$ are alternate.Now, by clearing the denominator in $f(x)$, we have $$f(x) = \sum\limits_{i=1}^{2020} f_i(x).$$ This shows that $f(x)$ is a polynomial of degree $2019$. From the above discussions, we observe that, for $1 \le i \le 2020$, $$f(i) = f_i(i).$$

This shows that, the function $f$ changes its sign when we go from $1$ to $2020$ by one integer to next integer. Hence, the function $f(x)$ has atleast one real zero in each of the interval $(i,i+1)$ ($1\le i \le2019$) and in total it has atleast 2019 real zeros. But $f(x)$ itself is a polynomial of degree $2019$ and hence this should be the full set of real roots of $f(x)$.

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