Let $\mathbb R$ be the set of real numbers. Determine all functions $f:\mathbb R\to\mathbb R$ that satisfy the equation\[ f(f(x))+yf(xy+1) = f(x-f(y)) + xf(y)^2. \]for all real numbers $x$ and $y$.
Problem
Source: Latvian TST for Baltic Way 2022 P3
Tags: functional equation, TST
Kimchiks926
26.10.2022 01:38
This problem was proposed by me.
Let $P(x;y)$ denote the assertion of the given functional equation. Observe that from $P(0;y)$ we have:
$$ f(f(0))+yf(1) = f(-f(y)) \qquad (1) $$Assume that $f(1) \neq 0$. In this case, if $f(a)=f(b)$, we have that:
$$ af(1) = f(-f(a))-f(f(0))=f(-f(b))-f(f(0)) = bf(1) \implies a=b $$This means that function $f$ is an injective function. Setting $y=0$ in the equation $(1)$ reveals us that:
$$ f(f(0)) =f(-f(0)) \implies f(0)=-f(0) \implies f(0) =0 $$Finally, considering $P(x,0)$ allows us to conclude that:
$$ f(f(x)) =f(x) \implies f(x) = x $$It is easy to check that this function does not work.
Now we focus on the case $f(1)=0$. Note that $P(0;1)$ gives us:
$$ f(f(0)) =f(-f(1)) =f(0) $$On the other hand, from $P(x;0)$ we have:
$$ f(f(x)) = f(x-f(0))+xf(0)^2 \qquad (2) $$Setting $x=f(0)$ in the last equation leads us to the following:
\begin{align*}
f(f(f(0)))=f(f(0)-f(0)) +f(0)^3 \\
f(f(0)) = f(0) +f(0)^3 \\
f(0) = 0
\end{align*}This means that equation $(2)$ can be transformed as $f(f(x))=f(x)$. To finish, we just need to consider $P(x;1)$:
$$ f(f(x))+f(x+1) = f(x) \implies f(x+1) = 0 $$It is easy to check that this function indeed works.
MathLuis
26.10.2022 04:43
Time to double down! If $f=0$ we are done so assume $f$ is non-zero (i.e. exists $d$ such that $f(d) \ne 0$), let $P(x,y)$ the assertion of the given F.E. $P(0,x)$ $$xf(1)+f(f(0))=f(-f(x)) \implies f \; \text{bijective} \; \text{or} \; f(1)=0$$Case 1: $f$ bijective $P(0,0)$ $$f(f(0))=f(-f(0)) \implies f(0)=-f(0) \implies f(0)=0$$$P(x,0)$ $$f(f(x))=f(x) \implies f(x)=x \; \text{contradiction!!}$$Case 2: $f(1)=0$ $P(0,1)$ $$f(f(0))=f(0)$$$P(x,0)-P(x,f(0))$ $$f(0)f(xf(0)+1)=0 \implies f(0)=0 \; \text{or} \; f(x)=0 \; \forall x \ne 0$$The later one is not possible becuase then $f(0)=f(f(0))=0$ so we get $f(0)=0$ $P(0,x)$ $$f(-f(x))=0$$$P(x,0)-P(x,-f(d))$ $$f(1-xf(d))=0 \implies f(t)=0 \; \text{contradiction!!}$$Hence the only solution that works its $f(x)=0$ thus we are done
megarnie
26.10.2022 14:30
The only solution is $\boxed{f\equiv 0}$, which works. Let $P(x,y)$ be the given assertion. Claim: If $f(1)\ne 0$, then $f$ is injective. Proof: Suppose $f(1) \ne 0$ and $f(a) = f(b)$. Comparing $P(0,a)$ and $P(0,b)$ gives $af(1) = bf(1)$, so $a=b$. $\square$ Case 1: $f(1) = 0$. $P(x,1): f(f(x)) + f(x+1) = f(x)$. Setting $x=0$ here gives $f(f(0)) = f(0)$. $P(f(0),0): f^3(0) = f(0) + f(0)^3$. However,\[f^3(0) = f(f^2(0)) = f(f(0)) =f(0),\]so $f(0)^3 = 0\implies f(0) = 0$. $P(x,0): f(f(x)) = f(x)$. Using $P(x,1)$ and $P(x,0)$, we get $f(x+1)= 0$, so $f\equiv 0$. Case 2: $f(1) \ne 0$. Then $f$ is injective. $P(0,0): f(f(0)) = f(-f(0))$, so $f(0) = 0$. $P(x,0): f(f(x)) = f(x)$. $P(0,x):f(-f(x)) = xf(1)$, so $f$ is surjective. If $f(k) = -1$, then $f(1) = kf(1)$, so $k=1$. Thus $f(1) = -1$. However, this gives $f(f(1)) = f(1)$, so $f(-1) = -1$, contradiction as $f$ is injective.
ZETA_in_olympiad
12.11.2022 11:50
Let $P(x,y)$ denote $f(f(x))+yf(xy+1)=f(x-f(y))+xf(y)^2$. $P(0,x)$ implies $f$ is bijective if $f(1)\neq 0$. Assume so, then $P(0,0)$ implies $f(0)=0$ and $P(x,0)$ implies $f(x)=x$ which fails. So $f(1)=0$ and $P(0,1)$ implies $f(f(0))=f(0)$. $P(f(0),0)$ implies $f(0)=0$ and $P(x,0)$ implies $f(f(x))=f(x)$. Finally, $P(x-1,1)$ implies $f(x)=0$, which works.