$Problem 16$
Find all function such that $f:R \rightarrow R$ and
$f(xf(y)+y)+f(xy+x)=f(x+y)+2xy$
$\textbf{Solution 16}$
Let $P(x,y)$ be the assertion.
$P(1,\frac{x}{2})\Rightarrow f(f(\frac{x}{2})+\frac{x}{2})=x \Rightarrow f$ is surjective $\Rightarrow$ there exists $a$ such that $f(a)=1$.
$P(\frac{x}{a+1},a)\Rightarrow f(x)=x\frac{2a}{a+1}$
Now, obviously $a \neq -1$, because assuming that $f(-1)=1,P(x,-1)$ gives $0=-2x$ which is not possible.
So, plugging $f(x)=cx$ in the main equation we have $f(x)=x,f(x)=-2x.\blacksquare$
Re: FE Marathon!
Posted: Sat Feb 27, 2021 7:02 pm
by ~Aurn0b~
$\textbf{Problem 17}$
Find all functions $f: \mathbb R \to \mathbb R$ such that\[ f( xf(x) + f(y) ) = f(x)^2 + y \]for all $x,y\in \mathbb R$.
$P(x,y)=f(x(f(x)+f(y)=f^2(x)+y$
$P(0,y)=f((y))=f^2(0)+y$
Let, $f(x_1)=f(x_2)$
Or,$f(f(x_1))=f(f(x_2))$
Or,$f^2(0)+x_1=f^2(0)+x_2$
$\therefore x_1=x_2$
Which means $f$ is one-one.
Now,
$P(x,0)=f(x(f(x)+f(0))=f^2(x)+0$
Or,$f(xf(x)+f(0))=f(f(x))$
Or,$xf(x)+f(0)=f(x)$
Implies, $f(x)=\frac{f(0)}{1-x} \forall x \in R$ and $x \neq 1$
The question changed so, this solution is no longer valid