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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:

  1. $f(-1)=-1$ (we can deduce it from $P(1,1),P(-1,1),P(-1,-2)$) and then conclude that $f(1)=1$
  2. $f(x+1)=f(x)+1$ (from $P(1,x)$)
  3. $f\left(x^2\right)=f(x)^2$
  4. $f(nx)=nf(x)$ for all integer $n$ and real $x$ (because $f(x)^2+\frac{f(nx)}{f(x)}=f\left(x^2+n\right)=f\left(x^2\right)+n$)
Mohsen Shahriari
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adem bizid
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  • Can you clarify how you arrived at statement 3? – Calvin Lin Dec 07 '21 at 00:08
  • Please avoid titles that only consist of Mathjax formulas. Always include some text. Also, including the key parts of the question in a compact form in the title is highly recommended. For example, "The functional equation $f\left(x^2+y\right)=f(x)^2+\frac{f(xy)}{f(x)}$ on $\mathbb R^*$" seems to be a good candidate for this post. – Mohsen Shahriari Dec 07 '21 at 01:58
  • @CalvinLin I suppose by letting $y=1$ in the original equation and using statement 2. – Mohsen Shahriari Dec 07 '21 at 02:00
  • @adembizid Calvin Lin's question is a good note on how a good post should be written down. I think it'd be good if you edited the post and added things like: 1. "where $\mathbb R^*=\mathbb R\setminus{0}$"; 2. "Let $P(x,y)$ denote the functional equation"; 3. "What I found is [number]. [what you've found] [how you've found it]"; 4. restrictions on the variables appearing in the equations (for example $x\ne0,-1$ in 2 or $n\ne x^2$ in the justification of 4, while adding an explanation for why it is still true that 4 holds for all $x$ and $n$) – Mohsen Shahriari Dec 07 '21 at 02:08

1 Answers1

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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.

jlammy
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