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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 $.

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    $\begingroup$ There are functions that are neither even nor odd. $\endgroup$
    – Arthur
    Commented Sep 28, 2019 at 17:33
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    $\begingroup$ The equation I got before I did the 2 cases implies that the function is either even or odd. $\endgroup$ Commented Sep 28, 2019 at 17:34
  • $\begingroup$ Looks right to me. $\endgroup$
    – saulspatz
    Commented Sep 28, 2019 at 18:09
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    $\begingroup$ When the product of two functions is $0$, that doesn't mean that one of the functions is the zero function. $\endgroup$
    – Arthur
    Commented Sep 28, 2019 at 18:20
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    $\begingroup$ 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. $\endgroup$
    – Arthur
    Commented Sep 28, 2019 at 19:07

2 Answers 2

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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.

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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 $.

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