Find all functions such that $f\left(x^2+y\right)=f(x)^2+\frac{f(xy)}{f(x)}$ in $\mathbb R^*$
Finds all function $f:\mathbb{R}^*\to\mathbb{R}^*$ such that $$\forall x,y\in\mathbb{R}^*,y\neq-x^2\qquad f\left(x^2+y\right)=f(x)^2+\frac{f(xy)}{f(x)}$$ where $\mathbb R^*=\mathbb R\setminus\{0\}$.
Let $P(x,y)$ denote the functional equation.
What I found is:
- $f(-1)=-1$ (we can deduce it from $P(1,1),P(-1,1),P(-1,-2)$) and then conclude that $f(1)=1$
- $f(x+1)=f(x)+1$ (from $P(1,x)$)
- $f\left(x^2\right)=f(x)^2$
- $f(nx)=nf(x)$ for all integer $n$ and real $x$ (because $f(x)^2+\frac{f(nx)}{f(x)}=f\left(x^2+n\right)=f\left(x^2\right)+n$)
Solution 1:
We know that for all $n\in\mathbb Z$, we have $f(n)=n$ and $f(x+n)=f(x)+n$ (using your properties 1 and 2). So plugging $x=b$, $y=a/b$ for $a,b\in\mathbb Z$ gives $$b^2+f(a/b)=f(b^2+a/b)=f(b)^2+f(a)/f(b)=b^2+a/b,$$ and hence $f(q)=q$ for all $q\in\mathbb Q$.
Next we claim that $f$ is strictly increasing. Property 3 implies that $f>0$ on $(0,\infty)$, so for $x,y>0$ we have $$f(x^2+y)=f(x)^2+f(xy)/f(x)>f(x^2),$$ so $f$ is increasing on $(0,\infty)$. We know that $f$ is odd (using property 4 with $n=-1$), so $f$ is increasing on $\mathbb R^*$.
So $f$ strictly increasing and $f(x)=x$ on $\mathbb Q$ together give that $f(x)\equiv x$ everywhere.