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Algebra Difficulty 5.7 AIME, harder Find the answer

Let R\mathbb{R} be the set of real numbers. Let f:RRf: \mathbb{R} \rightarrow \mathbb{R} be a function such that for all real numbers xx and yy, we have f(x2)+f(y2)=f(x+y)22xyf\left(x^{2}\right)+f\left(y^{2}\right)=f(x+y)^{2}-2 x y Let S=n=20192019f(n)S=\sum_{n=-2019}^{2019} f(n). Determine the number of possible values of SS.

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Solution

Letting y=xy=-x gives f(x2)+f(x2)=f(0)2+2x2f\left(x^{2}\right)+f\left(x^{2}\right)=f(0)^{2}+2 x^{2} for all xx. When x=0x=0 the equation above gives f(0)=0f(0)=0 or f(0)=2f(0)=2. If f(0)=2f(0)=2, then f(x)=x+2f(x)=x+2 for all nonegative xx, so the LHS becomes x2+y2+4x^{2}+y^{2}+4, and RHS becomes x2+y2+4x+4y+4x^{2}+y^{2}+4 x+4 y+4 for all x+y0x+y \geq 0, which cannot be equal to LHS if x+y>0x+y>0. If f(0)=0f(0)=0 then f(x)=xf(x)=x for all nonnegative xx. Moreover, letting y=0y=0 gives f(x2)=f(x)2f(x)=±xf\left(x^{2}\right)=f(x)^{2} \Rightarrow f(x)= \pm x for all xx. Since negative values are never used as inputs on the LHS and the output on the RHS is always squared, we may conclude that for all negative x,f(x)=xx, f(x)=x and f(x)=xf(x)=-x are both possible (and the values are independent). Therefore, the value of SS can be written as S=f(0)+(f(1)+f(1))+(f(2)+f(2))++(f(2019)+f(2019))=2i=12019iδiS=f(0)+(f(1)+f(-1))+(f(2)+f(-2))+\cdots+(f(2019)+f(-2019))=2 \sum_{i=1}^{2019} i \delta_{i} for δ1,δ2,,δ2019{0,1}\delta_{1}, \delta_{2}, \ldots, \delta_{2019} \in\{0,1\}. It is not difficult to see that S2\frac{S}{2} can take any integer value between 0 and 202020192=2039190\frac{2020 \cdot 2019}{2}=2039190 inclusive, so there are 2039191 possible values of SS.

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Source: Omni-MATH, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.