Problem from the 2019 Brazil Math Olympiad: $ f \big( x f ( y ) + f ( x ) \big) + f \left( y ^ 2 \right) = f ( x ) + y f ( x + y ) $

Question

The following functional equation problem:

Determine all the functions $ f : \mathbb R \to \mathbb R $ such that for any $ x , y \in \mathbb R $ it is true that: $$ f \big( x f ( y ) + f ( x ) \big) + f \left( y ^ 2 \right) = f ( x ) + y f ( x + y ) $$

This is the fourth problem from the university level paper here: https://www.obm.org.br/content/uploads/2019/11/Prova_Nivel_Universitario_OBM_2019.pdf

I am attaching an attempt in an answer (because would lengthen the description too much and which is good practice according to guidelines)

Answer 1

Edit: Fully Solved

This is my attempt at a solution and is partial (eventually resolves only in $\mathbb{Q}$):

Substituting $x=0$, $y=\beta$ and then $-\beta$, for $\beta$ any non zero real number, we get the equations: $$f(f(0)) + f(\beta^2) = f(0) + \beta f(\beta)\quad \quad \quad \quad \quad \quad \quad \quad.$$ $$f(f(0)) + f(\beta^2) = f(0) + -\beta f(-\beta)\quad \text{ subtracting, we get}$$ $$\implies f(-\beta) + f(\beta) = 0\quad \forall \beta \in \mathbb{R}-\{0\}\tag{1}$$

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Substituting $x=1$, $y=-1$, $$f(f(-1)+f(1)) + f(1) = f(1) - f(0) \quad \quad \text{ and now using (1) with }\beta=1\text{, we get:}$$ $$f(0) = -f(0) \implies f(0) = 0 \tag{2}$$

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Now with this knowledge if we substitute $x=0$, we have for any $y\in\mathbb{R}$: $$f(y^2) = yf(y) \tag{3}$$

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Also if we substitute $y=0$, with all this information, we obtain $\forall x\in \mathbb{R}$, $$f(f(x)) = f(x)\tag{4}$$

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Now substitute $x=-\beta, y=\beta$ and we get for all $\beta \in \mathbb{R}$: $$f(-\beta\cdot f(\beta)+f(-\beta)) + f(\beta^2) = f(-\beta)\quad \qquad \text{using (2)}$$ $$f((\beta+1)\cdot f(\beta)) = f(\beta^2) + f(\beta) = (\beta+1)\cdot f(\beta)\quad \text{using (1), (3)}$$ $$\text{Thus, }\qquad f((x+1)\cdot f(x)) = (x+1)\cdot f(x) \quad \forall x \in \mathbb{R}\tag{5}$$

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Now put $y=x$, and we get $\forall x\in\mathbb{R}-\{0\}$ (also for $0$, but that's trivial), $$f((x+1)\cdot f(x)) + f(x^2) = f(x) + xf(2x)$$ $$\implies (x+1)\cdot f(x) +xf(x) = f(x) + xf(2x)$$ $$\implies f(2x) = 2f(x) \tag{6}$$

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Earlier I went down the following route which resolves the problem in $\mathbb{Q}$. Feel free to skip ahead to the final solution section

Now you can establish by induction that this is true for all integer factors (not just $2$): briefly, if it is true for a factor of $k$ (say $2$), then in the inductive step, using assumption $f(kx) = > kf(x)$, we reach a relation for $k+1$. We substitute $x=ky$: $$f(kyf(y)+f(ky)) + yf(y) = f(ky) + yf((k+1)\cdot y)$$ $$\implies > f(k\cdot((y+1)\cdot f(y))) + yf(y) = kf(y) + yf((k+1)\cdot y)$$ $$\implies k\cdot((y+1)\cdot f(y)) + yf(y) = kf(y) + yf((k+1)\cdot y) > \qquad \text{using assumption and (5)}$$ $$\implies kyf(y) + kf(y) + > yf(y) = kf(y) + yf((k+1)\cdot y)$$ $$\implies f((k+1)\cdot y) = > (k+1)f(y)\qquad \text{ completing the induction}$$

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But this takes us nowhere in the real domain. We can resolve if the domain were $\mathbb{Q}$

Thus for rational numbers (negatives are easily determined from $(1)$ so consider positives for now), we would get, for $p, q\in > \mathbb{Z}^+$: $$f\bigg(\frac{p}{q}\bigg) = f\bigg(p\cdot > \frac{1}{q}\bigg) = p\cdot f\bigg(\frac{1}{q}\bigg)\qquad \text{ and } > \qquad f(1) = f\bigg(q\cdot\frac{1}{q}\bigg) = q\cdot > f\bigg(\frac{1}{q}\bigg)$$ $$\implies f\bigg(\frac{p}{q}\bigg) = > \frac{p}{q}\cdot f(1)$$ Thus assuming $f(1)=c$, for rationals our solution would be $f(x) = cx\quad\forall x\in \mathbb{Q}$ and substituting we get possible values of $c$ as $0$ and $1$, yielding two solutions: $$f(x)=0 \quad\forall x\in \mathbb{Q}\qquad \text{ and > } f(x)=x\quad\forall x\in \mathbb{Q}$$

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Final Solution

(Ref @1-___-'s comment) We can observe that this is a Cauchy functional equation by adding up the $x$ and $-x$ substitutions and use $(3)$ and $(6)$ to get: $$f(x+y) + f(y-x) = f(2y)$$ Since we can independently pick any real pair of values for $x+y$ and $y-x$, we have that $$f(\alpha)+f(\beta)=f(\alpha+\beta)\quad \forall\alpha,\beta\in\mathbb{R}$$

Now we can use this to target the original relation which now simplifies to: $$f(xf(y)) + f(f(x)) + f(y^2) = f(x) + yf(x) + yf(y)$$ Using $(3)$ and $(4)$, $$f(xf(y)) = yf(x)$$ Now substituting $x=1$, we get for all real $y$: $$f(f(y)) = f(y) = y\cdot f(1)$$ Thus the solution is in the space of linear functions: $$f(x) = c\cdot x$$ for some $c\in\mathbb{R}$. Plugging this form back into the original equation forces $c=0$ or $1$ and thus yields these two solutions in $\mathbb{R}$: $$f(x)=0 \quad\forall x\in \mathbb{R}\qquad \text{ and } f(x)=x\quad\forall x\in \mathbb{R}$$

Answer 2

Here's an alternative approach.

Clearly $f(x)=0$ satisfies the given equation.Let $f(x)$ not be $0$ everywhere.

$ Let \ P(x,y) \ denote\ the\ statement:\ \\ f \big( x f ( y ) + f ( x ) \big) + f \left( y ^ 2 \right) = f ( x ) + y f ( x + y )$

Then, plugging in $x=0 \ and \ y=0 $ gives

$ P(0,0) \Rightarrow f(f(0))=0 \tag{1} \\ P(0,f(0))\Rightarrow f(f(0))+f(f(0)^2)=f(0)+f(0)f(f(0)) \\$ $\qquad f(f(0)^2)=f(0) \tag{2}$

Now,plugging in $x=f(0)\ and\ y=0$ and using $(1)$ and $(2)$ we get

$$ f\big (\ f(0)^2+f(f(0)) \ \big) +f \big(\ 0 \ \big)= f(f(0)) + 0*f(\ f(0)\ ) \\ \implies f\big (\ f(0)^2 \ \big) +f \big(\ 0 \ \big)=0 $$ $\implies f(0)=0 \tag{3}$

Plugging in $x=0$ first and then $y=0$ after in our original equation, we get:

$$ P(x,0) \implies f \big(\ f(x)\ \big)=f(x) \tag{4} $$

$$ P(0,y) \implies f(y^2)=yf(y) \tag{5} $$ Now we try to investigate at which points does $f(x)$ vanish.We propose that it vanishes only when $x=0.$ $ Lemma \ 1: f(k)=0 \implies k=0.\\ Proof: P(x,k) \implies f(f(x))+f(k^2)=f(x)+ kf(x+k) \\ \implies f(x) +kf(k)=f(x) +kf(x+k) \qquad using\ (4)\ and\ (5) \\ \implies 0=kf(x+k) \implies k=0 \ as \ f(x) \ isn't \ identically \ 0.\\$

$ \\ Lemma \ 2: f(x) \ is\ an\ odd\ function.\\ Proof:\ Note\ that\ if \ x=0,\ then\ f(x)=-f(-x)\\ Now\ assume\ that\ x \neq 0.\ Then\ using\ (5)\\ f(x^2)=xf(x)=-xf(-x)\\ \implies f(x)=-f(-x)\\ $

As $(4)$ suggests, $f(x)$ should be an identity function.But we haven't proved it yet.We have only proved that f maps a real number to itself if that number belongs to the image(or range) of $f$.Let $I$ denote the set of all values which $f(x)$ can take. Then $\forall k \in\ I\ , \ f(k)=k.$ Note that if $k\in I$,then $-k\in I$ as $f$ is an odd function.After a thorough investigation of the elements in set $I$, we arrive at the following proposition:

$ Proposition: If \ \ k \in I \ and \ a \in \Bbb R \ such\ that\ f(a)=k .\ Then\ the\ following\ hold: \\ \bullet k \in I \\ \bullet -k \in I \\ \bullet k^2 \in I \\ \bullet k(1+a) \in I \\ \bullet k(1-a) \in I \\ \bullet k+a \in I \\ \bullet k-a \in I \\ Proof:\ f(a)=k \ , f(-a)=-k, \ f(k^2)= kf(k)=k^2 \\ P(a,-a) \implies f(k(1-a))+f(a^2)=k-af(0) \\ \implies f(k(1-a)) +af(a)=k \\ \implies f(k(1-a))=k(1-a)\\ Hence\ k(1-a) \in I\\ as\ k(1-a) \in I,\ so\ (-k)(1-(-a))\in I\ (as\ f(-a)=-k)\\ \implies (-k)(1+a)\in I\\ \implies\ k(1+a)\in I\\ P(a,k)\implies f(k(1+a))+f(k^2)=k+kf(a+k)\\ \implies k(1+a)+k^2=k +kf(a+k),\ as \ (k(1+a),k^2\in I)\\ \implies kf(a+k)=k^2+ka $

Now if $k=0$ then $a=0$, so both $k+a \in I$ and $k-a \in I$ holds.So let's assume that $k\neq 0$.Then $f(a+k)=k+a\\ Hence,\ k+a\in I\\ P(a,-k) \implies f(k(1-a))+f(k^2)=k-kf(a-k)\\ \implies k(1-a)+k^2=k+kf(k-a)\\ \implies f(k-a)=k-a\\ Hence\ k-a\in I.\\ Hence\ our\ proposition\ is\ true. $

Note that from our proposition we can deduce that $-f(1)\in I \ and\ f(1)-1\in I$.Subtituting them in our original equation, we get: $$ P(-f(1),f(1)-1)\implies f\big(\ -f(1)(f(1)-1)-f(1)\ \big)+f((f(1)-1)^2)=-f(1)+(f(1)-1)f(-1)\\ f(-f(1)^2)+(f(1)-1)^2=-f(1)-f(1)^2+f(1)\\ (as\ f(-1)=-f(1) \ and \ (f(1)-1)^2 \in I \ according\ to\ our\ proposition.)\\ Note\ that\ as\ f(1)^2 \in I ,so\ -f(1)^2 \in I.\ Hence\\ -f(1)^2 +(f(1)-1)^2=-f(1)^2\\ \implies f(1)=1\\ so f(-1)=-1\ $$

Note that $f(x)-x$ should be zero.So in order to get that expression we subtitute $y=-1$ in our original equation.So

$$ P(x,-1)\implies f(f(x)-x)+1=f(x)-f(x-1)\\ as\ f(x)-x\in I\ so\ we\ get,\\ f(x)-x+1=f(x)-f(x-1)\\ \implies f(x-1)=x-1\\ \implies f(x)=x $$

Therefore $f(x)=x \ \ \forall x\in \Bbb R$ or $ f(x)=0 \ \ \forall x\in \Bbb R$