# Functional equation problem: $f \left( y ^ 2 - f ( x ) \right) = y f ( x ) ^ 2 + f \left( x ^ 2 y + y \right)$

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 $$f ( x ) = 0$$.

• There are functions that are neither even nor odd. Sep 28, 2019 at 17:33
• The equation I got before I did the 2 cases implies that the function is either even or odd. Sep 28, 2019 at 17:34
• Looks right to me. Sep 28, 2019 at 18:09
• When the product of two functions is $0$, that doesn't mean that one of the functions is the zero function. Sep 28, 2019 at 18:20
• For any $x$ one of the brackets must be $0$. It doesn't have to be the same bracket each time. $f(x)=\sin(x)$ for $x\geq0$, $f(x)=|\sin(x)|$ for $x<0$ fulfills that property, but is neither even nor odd. Sep 28, 2019 at 19:07

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.
You can show that the only function $$f : \mathbb R \to \mathbb R$$ satisfying $$f \left( y ^ 2 - f ( x ) \right) = y f ( x ) ^ 2 + f \left( x ^ 2 y + y \right) \tag 0 \label 0$$ for all $$x , y \in \mathbb R$$ is the constant zero function, by continuing your own argument. You've established showing $$f ( - x ) ^ 2 = f ( x ) ^ 2 \text . \tag 1 \label 1$$ Now, take the "$$y f ( x ) ^ 2$$" term in \eqref{0} to the left-hand side, square both sides and rearrange the terms to get $$f \left( y ^ 2 - f ( x ) \right) ^ 2 + y ^ 2 f ( x ) ^ 4 - f \left( x ^ 2 y + y \right) ^ 2 = 2 y f ( x ) ^ 2 f \left( y ^ 2 - f ( x ) \right) \text . \tag 2 \label 2$$ Substituting $$- y$$ for $$y$$ in \eqref{2} and using \eqref{1}, the left-hand side will be equal to the left-hand side of \eqref{2}, while the right-had side will be the opposite of the right-hand side of \eqref{2}. This in particular implies that $$y f ( x ) ^ 2 f \left( y ^ 2 - f ( x ) \right) = 0 \text . \tag 3 \label 3$$ Now, if for some $$a \in \mathbb R$$ we have $$f ( a ) \ne 0$$, then setting $$x = a$$ and considering all nonzero values for $$y$$ in \eqref{3}, we can see that for all $$z > - f ( a )$$ we have $$f ( z ) = 0$$. But then letting $$x = a$$ and $$y = 1 + \frac { | f ( a ) | } { 1 + a ^ 2 }$$ in \eqref{0} leads to a contradiction with $$f ( a ) \ne 0$$, since we have $$y > 0$$, $$y ^ 2 - f ( a ) > - f ( a )$$ and $$a ^ 2 y + y > - f ( a )$$. Therefore $$f ( x )$$ must be equal to $$0$$ for all $$x \in \mathbb R$$.