Functional Equations revisited.

Note that (putting $x=y=0$ ) $2f(0) = f(0)^2$ , so that either $f(0)=0$ or $f(0)=2$ .

If $f(0)=2$ then $2f(x^2) \; = \ f(x^2) + f((-x)^2) \; =\; f(0)^2 + 2x^2 \; = \; 4 + 2x^2$ so that $f(u) = u + 2$ for all $u \ge 0$ . Then $x^2 + y^2 + 4 \; = \; f(x^2) + f(y^2) \;= \; f(x+y)^2 - 2xy$ so that $(x+y)^2 + 4 \; = \; f(x+y)^2 \; = \; (x+y+2)^2 \; = \; (x+y)^2 + 4 + 4(x+y)$ for any $x,y$ such that $x+y \ge 0$ . This is not possible, and hence $f(0)=0$ .

Thus $2f(x^2) \; = \; f(x^2) + f((-x)^2) \; = \; f(0)^2 + 2x^2 \; = \; 2x^2$ so that $f(u) = u$ for all $u \ge 0$ , and $x^2 + y^2 \; = \; f(x^2) + f(y^2) \; = \; f(x+y)^2 - 2xy$ so that $f(x+y)^2 \; = \; (x+y)^2$ for all $x,y$ , and hence $|f(u)| = |u|$ for all $u \in \mathbb{R}$ No other restrictions are placed on $f$ by the given condition. Thus we deduce that $S_f \; = \; \sum_{n=-2019}^{2019}f(n) \; = \; \sum_{n=1}^{2019}\big[f(n) + f(-n)\big] \; = \; \sum_{n =1}^{2019} \big[n + f(-n)\big] \; = \; 2\sum_{n \in A_f}n$ where $A_f \; = \; \big\{ n \in \mathbb{Z} \,\big|\,-2019 \le n \le -1 \;,\; f(-n) > 0 \big\}$ Thus $S_f$ can be any even integer between $0$ and $2 \times \tfrac12 \times 2019 \times 2020 = 2019 \times 2020$ . Thus there are $1 + 2019 \times 1010 = \boxed{2039191}$ possible values of $S_f$ .