Notice that $f$ must not be constant We prove following result: $f(0)=0$ $f(x^2)=f(x)^2$ $f(1)=1$ $f(x+1)=f(x)+1$ $f(x)=0 \Leftrightarrow x=0$, therefore $f$ is strictly increasing Now we prove that $f(x)=x$ for any positive rational $x$, and because $f$ is strictly increasing, we have $f(x)=x$ for all positive x. A little bit easy for Bulgaria TST, I think.

I think there is something wrong with your solution, Nguyenhuyhoang, $f$ is a function from the nonzero to the nonzero, so we cannot have $f(0)=0$.

Here's my solution: $f(x^2+y)=f^2(x)+\frac{f(xy)}{f(x)}$ (*) Substitute $y$ by $1$, we have: $f(x^2+1)=f^2(x)+1$, therefore $f(x) > 1 \forall x > 1$ (1), and for any nonzero $x$, either $f(x)=f(-x)$ or $f(x)=-f(-x)$. Substitute $x$ by $1$ and let $f(1)=a$, we have: $f(y+1)=a^2+\frac{f(y)}{a} \forall y \neq 0$. (2) Consider the case $a^2 \neq 1$: For an arbitrary $x_0$, ($x_0 \neq 0; x_0 \neq 1$), we consider two cases: *If $f(x_0)=f(-x_0)$ From (*), substitute $x$ by $x_0$, $y$ by $-1$, we have: $f(x_0^2-1)=f^2(x_0)+1=f(x_0^2+1)$ Using (2) we shall end up at: $f(x_0^2-1)=f(x_0^2+1)=f(x_0^2)=\frac{a^3}{a-1}$ Hence, $f(x)=\frac{a^3}{a-1} \forall x$ such that $x-x_0^2$ is an integer. Now we know that if we choose a number $x_0$ such that $x_0^4 = x_0^2+k$ ($k>0$, $k$ is an integer), we shall have $f(x_0^4+1)=f(x_0^2) \Leftrightarrow f^2(x_0^2)+1=f(x_0^2)$ (a contradiction). *If $f(x_0)=-f(-x_0) \forall x \neq 0$: With the similar technique, we shall end up at $f(x_0^2-1)=\frac{a^3+2a^2}{a+1} \forall x_0$ Hence $f(x)$ is constant for all $x>-1$ (a contradiction). So $a^2=1$. Consider the case $a=-1$: From (2) we deduce that $f(x+1)=1-f(x) \forall x \neq 0$ Choose $x>1$, we shall have $f(x)>1 \Rightarrow f(x+1) <1$ (a contradiction due to (1) ). So $a=1$. Now we can prove the problem with steps mentioned by Nguyenhuyhoang.

Oops, my bad, I misread the problem. I thought it was non-negative reals. Well, the key is to prove $f(1)=1$, and your solution is correct.

How about this one (just change the equation) $ f(x^2+y)=f(x^2)+\frac{f(xy)}{f(x)} $

I have no idea whether or not this solution is correct, so please PM me if this has any flaws. Denote $P(x,y)$, where $-x^2 \not= y$ as the assertion that $f(x^2+y)=f(x)^2+\frac{f(xy)}{f(x)}$. Note that we don't need to care about $-x^2 \not= y$ when $y >0$ and the denominator being $0$ and stuff like that. For some $x \not= 1$, using $P(x,1)$ and $P(-x,1)$, we have $f(x)^2+1=f(x^2+1)=f(-x)^2+1$, so $f(x)=f(-x)$ or $f(x)=-f(-x)$. Assume $f(x)=f(-x)$. I will show that $f(a)=f(-a)$ holds for all $a \in \mathbb{R}^{*}$. Use $P(x,y)$ and $P(-x,y)$, where $y \not= -x^2$. This gives $f(x)^2+\frac{f(xy)}{f(x)}=f(x^2+y)=f(-x)^2+\frac{f(-xy)}{f(-x)}$, and using $f(x)=f(-x)$, we get $f(xy)=f(-xy)$. Since $xy$ is surjective over $\mathbb{R}^{*}$, we have $f(a)=f(-a)$ for all $a \not= -y^2$. Now for some $t \not= -y^2$, we have $-t^2 \not= -y^2$, so since we have $f(t)=f(-t)$, we have $f(-y^2)=f(y^2)$ as well. This gives that $f(a)=f(-a)$ for all $a \in \mathbb{R}^{*}$. Similarly, if we have $f(x)=-f(-x)$, we have $f(a)=-f(-a)$ for all $a \in \mathbb{R}^{*}$. Set $f(1)=a$. $P(1,y)$ gives $f(y+1)=a^2+\frac{f(y)}{a}$. First, assume that $f(x)=f(-x)$. Now we have $f(x^2+1)=f(x)^2+1=f(x^2-1)$ for $x>1$. Now using $f(y+1)=a^2+\frac{f(y)}{a}$, we have $f(x^2-1)=f(x^2)=f(x^2+1)=\frac{a^3}{a-1}$. This gives us that $f(x)$ is a constant for sufficiently large $x$. Immediate contradiction as $a=a^2+1$ has no solutions. Therefore, we have $f(x)=-f(-x)$ for all $x$. Now for sufficiently large $x$, we have $f(x^2+1)=f(x)^2+1$ and $f(x^2-1)=f(x)^2-1$. Now using $f(y+1)=a^2+\frac{f(y)}{a}$ and solving for $f(x)^2$, we get $a^4+a^3-a^2-1=f(x)^2(a^2-1)$. Assume $a^2 \not= 1$. This enables us to claim that $f(x)^2$ is a constant for large enough $x$. This gives us that $f(x^2+1)$ is constant for large enough $x$, so $f(x)$ is constant for large enough $x$, contradiction. Now $a^2=1$. Also, $a=-1$ quickly fails as we need $a^4+a^3-a^2-1=0$. Therefore, we have $f(1)=1$. Now we have $f(y+1)=f(y)+1$ for $y \not= -1, 0$ and $f(x)=-f(-x)$, so $f(n)=n$ for all $n \in \mathbb{Z}$. We also have $f(x^2+1)=f(x^2)+1=f(x)^2+1$, giving $f(x^2)=f(x)^2$. Also, note that $f(x^2+1)=f(x)^2+1 >1$ for all $x$, so $f(x)>1$ if $x>1$. Now we have for $x,y>0$, $f(x^2+y)=f(x^2)+\frac{f(xy)}{f(x)} > f(x^2)$, so $f$ is strictly increasing since $f$ is odd. Summing $P(x,y)$ and $P(x,-y)$ gives $f(x^2+y)+f(x^2-y)=2f(x^2)$, so Cauchy FE holds. Now it is easy to show that $f(x)=x$ for all rationals $x$ and we can conclude by strictly increasing condition. We are done. The only answer is $\boxed{f(x) \equiv x \text{ } \forall x \in \mathbb{R}^{*}}$

$f(x^2+y)+f(x^2-y)=2f(x^2)$ This is not Cauchy equation .

For convenience, $(f(x))^2=f(x)^2$. Let $P(x,y)$ be the assertion, $$f(x^{2}+y) = f(x)^{2} + \frac{f(xy)}{f(x)}\ \forall x,y \in \mathbb{R}^{*} \text{ and} -x^{2} \neq y$$ Assume $f(x)$ is a constant function, $$f(x)=c, \implies c^2-c+1=0 \implies c \in \{e^{\frac{2\pi i}{3}}, e^{\frac{4\pi i}3}\} \not\subseteq \mathbb{R^*}$$ Now considering it to be a non-constant function, Claim 1: $f(1)=1$ Proof: Let $k=f(1)$. \begin{align*} P(1,1) &&\implies f(2) &=k^2+1\\ P(2,1) &&\implies f(5) &= f(2)^2+1 = k^4+2k^2+2\\ P(1,2) &&\implies f(3) &= k^2+\frac{f(2)}k = k^2+k+\frac1k\\ P(1,3) &&\implies f(4) &= k^2+\frac{f(3)}k = k^2+k+1+\frac1{k^2}\\ P(1,4) &&\implies f(5) &= k^2+\frac{f(4)}k = k^2+k+1+\frac1k+\frac1{k^3}\\ \end{align*}$$\implies k^2+k+1+\frac1k+\frac1{k^3} = f(5) = k^4+2k^2+2$$$$\iff \frac{k^7+k^5-k^4+k^3-k^2-1}{k^3}=\frac{(k-1)(k^2-k+1)(k^2+k+1)^2}{k^3}=0$$$$\implies f(1) = k = 1$$ Claim 2: $f(x+n)=f(x)+n \ \forall x\in \mathbb{R^*} \text{ and } n \in \mathbb{N}$ Proof: $P(1,x)$ $$f(1+x)=k^2+\frac{f(x)}{k}=f(x)+1$$$$f(x+n)=f(x+n-1)+1=f(x+n-2)+2 = ... = f(x)+n\ \forall n \in \mathbb{N}$$ Claim 3: $f(n)=n \ \forall n \in \mathbb{N}$ $$\because f(x+n)=f(x)+n \implies f(n+1) = n + 1 \implies f(n) = n \ \forall n \in \mathbb{N}_{\geq 2}$$$$\text{But since, } f(1) = 1, f(n)=n\ \forall n \in \mathbb{N}$$ Claim 4: $f(n)=n\ \forall n \in \mathbb{Z}^*$ Proof: $$f(x+n)=f(x)+n\ \forall n \in \mathbb{N}$$Replacing $x$ with $1-n$, $$1=f(1)=f(1-n+n)=f(1-n)+n$$$$\implies f(1-n)=1-n$$Since $f(0)$ is not defined, $n \neq 1$. $$\implies f(x)=x\ \forall x \in \mathbb{Z^*}$$ Claim 5: $f(nx)=nf(x) \ \forall x \in \mathbb{R^*}, n \in \mathbb{Q^*}$ Proof: $$f(x+n)=f(x)+n\ \forall x \in \mathbb{R^*}, n\in \mathbb{N}$$$$\implies f(x+n^2)=f(x)+n^2 \ \forall x\in \mathbb{R^*}, n^2 \in \mathbb{N}$$$P(n, x), x\in \mathbb{R^*}, n \in \mathbb{N}$ $$f(x+n^2)=f(n)^2+\frac{f(nx)}{f(n)}=n^2+\frac{f(nx)}{n}$$$$\implies n^2+f(x)=n^2+\frac{f(nx)}{n} \implies nf(x)=f(nx) \ \forall n \in \mathbb{N}, x \in \mathbb{R^*}$$ Claim 6: $f(x)=-f(-x) \ \forall x \in \mathbb{R}$ Proof: $P(-n,x), n\in \mathbb{N}, x\in\mathbb{R^*}$ $$f(n^2+x)=f(-n)^2+\frac{f(-nx)}{f(-n)}=(-n)^2+\frac{nf(-x)}{-n}$$$P(n,x) n\in \mathbb{N}, x\in\mathbb{R^*}$ $$f(n^2+x)=f(n)^2+\frac{(nx)}{f(n)}=n^2+\frac{nf(x)}{n}$$$$\implies n^2+\frac{nf(x)}{n} = f(n^2+x) =(-n)^2+\frac{nf(-x)}{-n} \implies -f(x)=f(-x)$$ Claim 7: $f(x)=x \ \forall x \in \mathbb{Q^*}$ Proof: $$nf(x)=f(nx) \ \forall n \in \mathbb{N}, x \in \mathbb{R^*}$$Setting $x=\frac1n$, $$nf\left(\frac1n\right) = f(1) = 1 \implies f\left(\frac1n\right) = \frac1n \ \forall n\in\mathbb{N}$$Let $\frac{p}q \in \mathbb{Q^*}$, such that $p \in \mathbb{Z^*}, q \in \mathbb{N}$. $$f\left(\frac{p}q\right) = f\left(p\frac1q\right) = pf\left(\frac1q\right) = \frac{p}q$$$$\implies f(x) = x\ \forall x \in \mathbb{Q^*}$$ Claim 8: $f(x^2) = f(x)^2 \ \forall x \in \mathbb{R^*}$ and $f(x) > 0 \ \forall x\in\mathbb{R}_{>0}$ Proof: As proved earlier, $$f(x^2+1) = f(x)^2+1$$Also as proved earlier, $$\because f(x+1)=f(x)+1\implies f(x^2+1)=f(x^2)+1$$Equating these gives $f(x^2)=f(x)^2 \forall x\in\mathbb{R^*}$. Also, $f(x^2) > 0 \implies f(x) > 0 \ \forall x \in \mathbb{R^+}$. Claim 9: $f(x^3)=f(x)^3 \ \forall x\in\mathbb{R^*}$ Proof: $$2f(x)^2=2f(x^2)=f(2x^2)$$$P(x,x^2)$ $$f(x^2+x^2)=f(x)^2+\frac{f(x^3)}{f(x)}\implies f(x^3)=f(x)^3$$$$\implies 2f(x)^2=f(x^2+x^2)=f(x)^2+\frac{f(x^3)}{f(x)}\implies f(x^3)=f(x)^3$$ Claim 10: $f(x)$ is monotonically increasing Proof: Let $x^2 > y > 0 \implies x^3> xy > 0$, $P(x^2,-y)$ $$f(x^2-y)=f(x)^2+\frac{f(-xy)}{f(x)}=f(x)^2-\frac{f(xy)}{f(x)}>0, \because x^2-y > 0$$$$\implies f(x^3)=f(x)^3>f(xy)\implies f(x)>f(y) \forall x > y > 0$$$$\because f(x)=-f(-x), f(x)<f(y)\ \forall 0<x<y$$$$\implies f(x) > f(y)\ \forall x>y \in \mathbb{R^*}$$ Claim 11: $f(x)=x \ \forall x \in \mathbb{R^*}$ Proof: Let $r_n$ and $R_n$ be be an increasing and decreasing sequence of rational numbers which converge towards $x \in\mathbb{R^*} - \mathbb{Q^*}$. $$r_n = f(r_n) \leq f(x) \leq f(R_n)=R_n$$As $n\to\infty$, $r_n$ and $R_n$ converge to $x$. $$\implies \boxed{f(x) = x\ \forall x \in \mathbb{R^*}}$$

Let $P(x,y)$ be the assertion of the problem statement.

As $f(x) = x$ for all rationals and $f$ is strictly increasing, \[\boxed{f(x) = x \text{ } \forall x\in \mathbb{R}^{*}}.\]

Can someone check if this is right? It's pretty much the same as above, but I got a slightly weaker increasing condition

Hence, given any real number $m>0$, we may take rationals $q_1$ and $q_0$ such that $q_1>m>q_0$. It follows,$$q_1=f(q_1)>f(m)>f(q_0)=q_0,$$Letting $q_1$ and $q_0$ approach $m$, we see that $f(m)=m$ since rationals are dense in reals. Hence, $f(x)=x$ for all $x>0$. But $f$ is odd, so $\boxed{f(x)=x}$ for all $x$.

$\textbf{\color{blue}{Answer:}}$ $f(x)=x$. $\textbf{\color{red}{Proof:}}$ Let $F(x,y)$ denote the given assertion, $\color{green}{\bullet f(1)=1}$ Comparing $F(1,1)$ and $F(-1,1)$ gives $f(2)=f(-1)^2+1$ and $f(1)^2=f(-1)^2 \implies f(1)=\pm f(-1)$. $F(-1,-2) \implies f(-1)=f(-1)^2+\frac{f(2)}{f(-1)}=f(-1)^2+\frac{f(-1)^2+1}{f(-1)} \implies f(-1)=-1$. If $f(1)=-1$, then $F(1,x^2)\implies f(1+x^2)=1-f(x^2)$ and $F(x,1)\implies f(x^2+1)=1+f(x)^2$ which means $f(x)^2=-f(x^2)\ge 0 $, so $f(x)\le 0$ when $x\ge 0$ but this contradicts with $f(1+x^2)=1+f(x)^2\ge 1$. So, $f(1)=1$. $\quad \square$ $\color{green}{\bullet f(n)=n \quad \forall n\in \mathbb{Z}}$ Since $f(1)=1$ and $f(-1)=-1$, this is evident from $F(1,x)\implies f(1+x)=1+f(x)$. $\quad \square$ $\color{green}{\bullet f(x)=x \quad \forall x\in \mathbb{Q}}$ Let $p,q\in \mathbb{Z}$, $F\left(q, \frac{p}{q}\right)\implies q^2+f\left(\frac{p}{q}\right)=f\left(q^2+\frac{p}{q}\right)=f(q)^2+\frac{f(p)}{f(q)}=q^2+\frac{p}{q} \implies f\left(\frac{p}{q}\right)=\frac{p}{q}. \quad \square$ $\color{green}{\bullet f(nx)=nf(x) \quad \forall n\in \mathbb{Z}}$ $F(n,x) \implies n^2+f(x)=f(n^2+x) = f(n)^2+\frac{f(nx)}{f(n)}=n^2+\frac{f(nx)}{n}\implies f(nx) = nf(x). \quad \square$ (This also shows that $f$ is odd by letting $n=-1$) $\color{green}{\bullet f(x)>0 \quad \forall x>0}$ $f(x)^2+1=f(x^2+1)=f(x^2)+1 \implies f(x^2)=f(x)^2\ge 0$. $\quad \square$ $\color{green}{\bullet f \textrm{ is strictly increasing}}$ $f(x^{2}+y) = (f(x))^{2} + \frac{f(xy)}{f(x)}>f(x)^2$ when $x,y>0$ and by the oddness of $f$, $f$ is indeed strictly increasing. $\quad \square$ Therefore, since $f$ is monotone with $f(x)=x$ for all rational $x$, $f\equiv x$. $\quad \blacksquare$

My solution from WOOT: Let $P(x,y)$ denote the given assertion. Claim: $f(1)=1$ Let $c=f(1)$. $P(1,1)\Rightarrow f(2)=c^2+1$ $P(1,2)\Rightarrow f(3)=c^2+c+\frac1c$ $P(1,3)\Rightarrow f(4)=c^2+c+1+\frac1{c^2}$ $P(1,4)\Rightarrow f(5)=c^2+c+1+\frac1c+\frac1{c^3}$ $P(2,1)\Rightarrow c^2+c+1+\frac1c+\frac1{c^3}=c^4+2c^2+2$ We obtain a $7$-th degree polynomial in $c$: $$c^7+c^5-c^4+c^3-c^2-1=0,$$which factors as $$(c-1)\left(c^2-c+1\right)\left(c^2+c+1\right)^2=0.$$Thus, $f(1)=1$. $P(1,x)\Rightarrow f(x+1)=f(x)+1$ $P(x,y+1)-P(x,y)\Rightarrow f(x)^2+\frac{f(xy)}{f(x)}+1=f(x)^2+\frac{f(xy+x)}{f(x)}\Rightarrow f(xy)+f(x)=f(xy+x)$ Taking $y\mapsto\frac yx$, we have $f(x+y)=f(x)+f(y)$. $P(x,1)\Rightarrow f(x^2)=f(x)^2\Rightarrow f(x)\ge0\forall x\ge0$ For $x\ge0$, we have $f(x+y)=f(x)+f(y)\ge f(y)$, hence $f$ is monotonically increasing. From additivity we obtain $f(x)=cx$ for some $c\in\mathbb R^*$. After testing we see that the only solution is $\boxed{f(x)=x}$.

I have a very complicated finish. Let $P(x,y)$ denote the given condition. $P(x,1)$ gives us $f(x^2+1)=f(x)^2+1$ for all $x$ and $P(1,x)$ implies $f(1+x)=f(1)^2+f(x)/f(1)$ for all $x\neq -1.$ Let $f(1)=t.$ Using the latter identities, we can compute $f(5)$ in two ways, as such \[\begin{cases}f(5)=f(2^2+1)=f(2)^2+1=(t^2+1)^2+1 \\ f(5)=t^2+f(4)/t=t^2+t+f(3)/t^2=\cdots=t^2+t+1+1/t+1/t^3\end{cases}\]It follows that $t^7+2t^5+2t^3=t^5+\cdots+t^2+1$ or, in other words, $t^3(t^4+t^2+1)=t^4+t^2+1.$ We infer that $f(1)=1.$ By substituting this in $P(1,x)$ we get $f(x+1)=f(x)+1$ for all $x\neq -1.$ Therefore, $f(n)=n$ for all $n\in\mathbb{N}.$ Moreover, by using $f(x+1)=f(x)+1$ in $P(1,x),$ we have $f(x^2)+1=f(x^2+1)=f(x)^2+1$ so $f(x^2)=f(x)^2$ for all $x\neq -1.$ Note that $f(x^2)=f(x)^2$ implies that $f(x)>0$ for all $x>0.$ Then, for $x,y>0$ we have \[f(x^2+y)=f(x)^2+f(xy)/f(x)=f(x^2)+f(xy)/f(x)>f(x^2)\]so $f$ is increasing on positive numbers. In particular, combining this with $f(x)>0$ for $x>0$ we have $0<f(r)<1$ for $0<r<1.$ Now, I can finish the solution like a normal person... or I can do this: Since $f(x+1)=f(x)+1$ for all $x\neq 1$ we know that $f(x)=\lfloor x\rfloor+f(\{x\})$ for all $x>0.$ We also know $f(x^2)=f(x)^2$ for all $x\neq -1$ (implicitely, for all $x>0$ as well). Therefore, for all $x>0$ we have\[(\lfloor x\rfloor+f(\{x\}))^2=f(x)^2=f(x^2)=\lfloor x^2\rfloor+f(\{x^2\}).\]For simplicity, let $\{x\}=\varepsilon_1, \ f(\{x\})=\varepsilon_2,$ and $f(\{x^2\})=\varepsilon_3.$Moreover, $\lfloor x\rfloor=t.$ We substitute to get \[t^2+2t\varepsilon_2+\varepsilon_2^2=(t+\varepsilon_2)^2=\lfloor(t+\varepsilon_1)^2\rfloor+\varepsilon_4=t^2+\lfloor2t \varepsilon_1+\varepsilon_1^2\rfloor+\varepsilon_3\]or, in other words, $\lfloor2t\varepsilon_1+\varepsilon_1^2\rfloor=2t\varepsilon_2+\varepsilon_2^2-\varepsilon_3.$ Using the fact that $\lfloor a\rfloor=b\iff b\leq a<b+1$ we can rewrite the latter as \[\varepsilon_2^2-\varepsilon_3-\varepsilon_1^2\leq 2t(\varepsilon_1-\varepsilon_2)\leq \varepsilon_2^2-\varepsilon_3-\varepsilon_1^2+1.\]but we know that $0<\varepsilon_1,\varepsilon_2,\varepsilon_3<1$ so basically, $2t(\varepsilon_1-\varepsilon_2)$ is bounded. Note that we can have $t\to\infty$ as $\varepsilon_1$ and $\varepsilon_2$ are fixed $($i.e. $\lfloor x\rfloor\to\infty$ as $\{x\}$ and $f(\{x\})$ are fixed$)$ so unless $\varepsilon_1=\varepsilon_2$ $($i.e. $f(\{x\})=\{x\})$ then as $t\to\infty$ we have $2t(\varepsilon_1-\varepsilon_2)\to\pm\infty$ which is a contradiction to our previous bounding. therefore, for all $x$ we have $f(\{x\})=\{x\}.$ Using $f(x)=\lfloor x\rfloor+f(\{x\})$ for $x>0$ we can infer that $f(x)=x$ for $x>0.$ After a couple simple substitutions, we easily infer that $f(x)=x$ for all $x.$

If I recall correctly, $f(x^2)=f(x)^2$ and $f(x+1)=f(x)+1$ over $\mathbb R^*$ is known.

What does $\forall$ mean?

falco_sparverius wrote: What does $\forall$ mean? As the LaTeX macro (\forall) suggests, it means for all something. So if you say $3\mid 2^n-1, \ \forall n=2k$ it means that $3$ divides $2^n-1$ for all $n$ of the form $2k.$

Nice one. Notice that $P(1,1),P(1,2),P(2,1),P(1,3),P(1,4)$ generates a 5 degree equation with variables $f(1),f(2),f(3),f(4),f(5)$. Solving it (I'm too lazy to do it) you get $f(1)=1$. Now: $f(y+1)=f(y)+1$, making $y=x^2$ we get: $f(x^2+1)=f(x^2)+1=f(x)^2+1$, where the last equality comes from $P(x,1)$ on the original equation. Hence $f(x^2)=f(x)^2 (I)$. Now let $y$ be $y+1$: $f(x^2+y+1)=f(x^2+y)+1=f(x)^2+f(xy+x)/f(x)$ Then notice: $f(x^2+y)+1=1+f(x)^2+f(xy)/f(x)$ by the original equation. Thus: $f(xy)+f(x)=f(xy+x)$, and taking $y=z/x$ we get that it is additive ($f(z)+f(x)=f(z+x)$), and it is bounded on positive reals by $(I)$, hence it is linear, and plugging we get that $f(x)=x$, q.e.d.