let $P(x,y)$ denote $f(x^2)-f(y^2)+2x+1=f(x+y)f(x-y)$ $P(0,0): f(0)^2=1$ if $f(0)=1$ $P(x/2,x/2)$$:f(x)=x+1$ wich is a solution if $f(0)=-1$ $P(x/2,x/2):$ $f(x)=-x-1$ wich is a solution aswell so the only solution are $:f(x)=x+1$ and $f(x)=-x-1$

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mathisreal wrote: Find all functions $R-->R$ such that: $f(x^2) - f(y^2) + 2x + 1 = f(x + y)f(x - y)$ I'm suprised this is the only (if i'm wrong) FE on Cono Sur Olympiad right?. Ok lets solve this in $5$ minutes lol. Let $P(x,y)$ the assertion of the given equation. $P(0,0)$ $$f(0)-f(0)+1=f(0)^2 \implies f(0)= \pm 1$$Now we kill the problem using $P \left(\frac{x}{2},\frac{x}{2} \right)$ $$x+1=f(x)f(0) \implies f(x)= \pm (x+1)$$And this fits so hence: $\boxed{f(x)=\pm (x+1) \; \forall x \in \mathbb R}$ Thus we are done

I think this Functional Equation is so well for begginers, btw it wasn't hard for be a Cono Sur.

My solution Let $P(x,y)$ denote the given assertion. $P(0,0):1=f(0)^2 \implies f(0)=\pm 1$ $P(\frac{x}{2},\frac{x}{2}):f(0)f(x)=x+1 \implies f(x)=\pm (x+1)$ if $f(x)=-(x+1)$, \begin{align*} f(x^2)-f(y^2)+2x+1=f(x+y)f(x-y) &\implies -(x^2+1)+(y^2+1)+2x+1 = (x+y+1)(x-y+1) \\ &\implies -x^2+y^2+2x+1=x^2-y^2+2x+1 \end{align*}, a contradiction! Therefore, the only solution is $\boxed{f(x)=x+1(\forall x \in \mathbb{R})} $, which clearly works $\blacksquare$