Is this function $f(x f(x)+y)=x^{2}+f(y)$ onto ??

Question

question -

Is this function $f(x f(x)+y)=x^{2}+f(y)$ onto $\mathbb R\to\mathbb R$?

I just need to know that is this function is onto or not ... by putting $y=0$ and $f(0)=s$ , I get $f(x f(x))=x^2 + s$ but i am not able to find an x such that $f(x)=y$ ...

any help will be appreciated

thankyou

Answer 1

With $a:=f(1)$, we have $$ f(a+y)=1+f(y)$$ hence by induction $$\tag1 f(na+y)=n+f(y)$$ for all $n\in\Bbb N$, and by letting $y\leftarrow y-na$ also $$ f(y)=n+f(y-na)$$ i.e., $(1)$ holds for all $n\in\Bbb Z$. In particular, $$ f(na)=n+f(0)\qquad\text{for }n\in\Bbb Z.$$ Given $z\in \Bbb R$, let $n=\lfloor z-f(0)\rfloor$ and $y=na$ so that $f(y)\le z$. Now let $x=\sqrt {z-f(y)}$ to find $$ f(xf(x)+y)=x^2+f(y)=z.$$