This functional equation problem is from the Latvian Baltic Way team selection competition 2019:
Find all functions $ f : \mathbb R \to \mathbb R $ such that for all real $ x $ and $ y $,
$$ f \left( y ^ 2 – f ( x ) \right) = y f ( x ) ^ 2 + f \left( x ^ 2 y + y \right) \text . \tag 1 \label {eqn1} $$
OK, so I think that the only answer is $ f ( x ) = 0 $.
I just want to see if my proof that it is the only solution is correct.
So we start off by plugging $ y = – y $. We get that
$$ f \left( y ^ 2 – f ( x ) \right) = – y f ( x ) ^ 2 + f \Big( – y \left( x ^ 2 + 1 \right) \Big) \text . $$
Then we add the two equations together getting that
$$ 2 f \left( y ^ 2 – f ( x ) \right) = f \Big( – y \left( x ^ 2 + 1 \right) \Big) + f \Big( y \left( x ^ 2 + 1 \right) \Big) \text . $$
From the above equations we get that
$$ \frac { f \Big( – y \left( x ^ 2 + 1 \right) \Big) + f \Big( y \left( x ^ 2 + 1 \right) \Big) } 2 = y f ( x ) ^ 2 + f \left( x ^ 2 y + y \right) \text. \tag 2 \label {eqn2} $$
Now, if we plug $ x = – x $ then we will get that the LHS is the same and that the RHS is $ y f ( – x ) ^ 2 + f \left( x ^ 2 y + y \right) $.
So we proceed by subtracting the two and getting that
$$ 0 = y f ( x ) ^ 2 – y f ( – x ) ^ 2 \text . $$
So, lets assume that $ y \ne 0 $ getting that
$$ 0 = \big( f ( x ) – f ( – x ) \big) \big( f ( x ) + f ( – x ) \big) \text . $$
Now we do a two case analysis, 1) the function is even and 2) the function is odd.
Lets start with the function being even then from \eqref{eqn2} we get that
$$ 0 = y f ( x ) ^ 2 \text , $$
which of course implies that the function is just $ 0 $.
OK, now the odd case. Since the function is odd, $ f ( 0 ) = 0 $. Then plugging $ x = 0 $ in \eqref{eqn1} we get that
$$ f \left( y ^ 2 \right) = f ( y ) \text , $$
which implies that $ f $ is also an even function. Since $ f $ is both even and odd, it can only be $ 0 $.
Since we got that $ f $ is zero in both cases, the only solution to the equation is $ f ( x ) = 0 $.
Best Answer
Suppose we have $(x^2+1)^2+4f(x)\geq 0$ for some $x$. Then there exists some $y_0$, such that $$y_0^2-(x^2+1)y_0-f(x)=0.$$ We may also assume that $y_0\ne 0$, because the roots of the above quadratic can't both be $0$. Plugging $P(x, y_0)$ in the equation we get $$y_0f(x)^2=0,$$ and because $y_0\ne 0$, we must have $f(x)=0$.
This directly implies $f(x)\leq 0$ for all $x$. Suppose the function does not positive roots. This means that we must have $f(x)<\frac{-(x^2+1)^2}{4}$ for all positive $x$. If we plug $P(x, 0)$ in the equation, where $x$ is positive, we get $$f(0)=f(-f(x))<-\frac{(f(x)^2+1)^2}{4}<-\frac{\left(-\left(\frac{(x^2+1)^2}{4}\right)^2+1\right)^2}{4},$$ where the first inequality follows from $-f(x)>0$ and $f(x)<\frac{-(x^2+1)^2}{4}$, whereas the second one follows from the fact that squaring the inequality mentioned above implies $$f(x)^2>\left(\frac{-(x^2+1)^2}{4}\right)^2.$$ Seeing as the $RHS$ of the inequality is not bounded from below, we have reached a contradiction, therefore we must have a positive root $a$. Plugging $P\left(x, \frac{a}{x^2+1}\right)$ in the equation we get $$0\geq f\left(\left(\frac{a}{x^2+1}\right)^2-f(x)\right)=\frac{a}{x^2+1}f(x)^2\geq 0,$$ so $\boxed{f\equiv 0}$, which indeed is a solution.