I don't know why you say this problem is not hard, but everybody has its own opinion. So, let's give a monstruous solution: First, it is trivial to prove that f(0)=0, f(1)=1. Suppose e managed to show that f(2)=2. Put y=2 and you wil find that $ f(x - 1) = \frac {f(x) + f(x - 2)}{f(x) - f(x - 2)}$ and an easy induction shows that f(n)=n for any natural n. Now we use the fact that $ \frac {f(x) + f(y)}{f(x) - f(y)} = f(\frac {x + y}{x - y}) = f(\frac {1 + \frac {y}{x}}{1 - \frac {y}{x}}) = \frac {1 + f(\frac {y}{x})}{1 - f(\frac {y}{x})}$. Then, it's immediately to show that $ f(x) = x$ for any rational x and that $ f(x)\geq 0$ for any nonnegative x. So, for any $ x\geq y\geq 0$ we have $ f(x) - f(y)\geq 0$ since this one has the same sign as $ f(\frac {x + y}{x - y})$. Thus, f is increasing on the positive and also multiplicative. It follows that f(x)=x for any positive x. It is also trivial to show that f is odd, so f(x)=x for any x. Now, the main thing: proving that f(2)=2. Put $ f(2) = a$. Then $ f(4) = a^2$ and also $ f(3) = f(\frac {4 + 2}{4 - 2}) = \frac {a + 1}{a - 1}$. Using that $ f(x + 1) = \frac {f(x) + f(x + 2}{f(x + 2) - f(x)}$ we find that $ f(5) = \frac {a^2 + 1}{(a - 1)^2}$. But $ f(5) = f(\frac {3 + 2}{3 - 2}) = ... = \frac {a^2 + 1}{2a + 1 - a^2}$. Thus, $ (a - 1)^2 = 2a + 1 - a^2$ from where a=0 or a=2. If a=0, then f(3)=-1 so $ f^2(\sqrt(3)) = - 1$ false.

I didn't mean that the problem was easy, I just meant that it wasn't hard to the extent that any of those parts is not obtainable. The (union of) the Singapore team got almost every part except showing that the function was increasing on positive (which, as we found out, is not difficult, it's just that we don't have enough experience with it). Whatever the case is there are quite a number of things to be shown. The official solution I have with all the details is 2 full pages long. Small note regarding the ending: It's easy to observe that the function is injective (if f(x) = f(y) but x <> y, then the LHS is defined but the RHS is undefined). Then since we have shown that f(0) = 0, a <> 0.

Valiowk wrote: Find all functions $ f: \mathbb{R} \to \mathbb{R}$ satisfying \[ f\left(\frac {x + y}{x - y}\right) = \frac {f\left(x\right) + f\left(y\right)}{f\left(x\right) - f\left(y\right)} \] for all $ x \neq y$. Harazi, how do you show that f(x) >= 0 for x >= 0 ?? Here is all I have (not a complete solution, just some ideas which could be of use for finding a simpler solution): We have given the equation $ \text{\textbf{(1)\ \ \ \ \ \ \ \ \ \ }}f\left( \frac{x+y}{x-y}\right) =\frac{f\left( x\right) +f\left( y\right) }{f\left( x\right) -f\left( y\right) }$. At first, our function f(x) must be injective; else, we would have x and y with f(x) = f(y) while $ x\neq y$, so that the right member of (1) would get undefined while the left member wouldn't. Putting y = 0 in (1), we get $ f\left( 1\right) =\frac{f\left( x\right) +f\left( 0\right) }{f\left( x\right) -f\left( 0\right) }$, what becomes (1 - f(1))f(x) = -f(0)(1 + f(1)). Now, if f(1) were not 1, we could divide by 1 - f(1), and f(x) would be a constant. This is not possible since f(x) is injective; hence, f(1) = 1. Therefore, -f(0)(1 + f(1)) = (1 - f(1))f(x) = 0. But since $ 1 + f\left( 1\right) = 2\neq 0$, we must have f(0) = 0. As a consequence, $ f\left( x\right) \neq 0$ for every $ x\neq 0$ (since f(x) is injective). Putting y = -x in (1), we get $ f\left( 0\right) =\frac{f\left( x\right) +f\left( -x\right) }{f\left( x\right) -f\left( -x\right) }$. Using f(0) = 0, this immediately becomes f(-x) = -f(x). Consequently, f(-1) = -f(1) = -1. Now let $ a\neq 0$ be a real number; then, $ \frac{f\left( x\right) +f\left( y\right) }{f\left( x\right) -f\left( y\right) }=f\left( \frac{x+y}{x-y}\right) =f\left( \frac{ax+ay}{ax-ay} \right) =\frac{f\left( ax\right) +f\left( ay\right) }{f\left( ax\right) -f\left( ay\right) }$, what simplifies to f(x)f(ay) = f(y)f(ax). With y = 1, this becomes f(x)f(a) = f(1)f(ax) = f(ax). (This trivially holds for x = 1 and for a = 0, even if these cases were not allowed in the proof.) Hence, we have f(x)f(a) = f(ax). In other words, our function f(x) is multiplicative. As a consequence, $ f\left( x\right) f\left( \frac{1}{x}\right) =f\left( 1\right) =1$ and $ \text{\textbf{(2)\ \ \ \ \ \ \ \ \ \ }}f\left( \frac{1}{x}\right) =\frac{1}{f\left( x\right) }$. Now, substitute y = 1 in (1). Then, with f(1) = 1 we obtain $ \text{\textbf{(3)}\ \ \ \ \ \ \ \ \ \ }f\left( \frac{x+1}{x-1}\right) =\frac{ f\left( x\right) +1}{f\left( x\right) -1}$. The quadratic equation $ \text{\textbf{(4)\ \ \ \ \ \ \ \ \ \ }}\frac{x+1}{x-1}=x$ has two real solutions: $ x = 1 + \sqrt {2}$ and $ x = 1 - \sqrt {2}$. But (3) shows that if x is a solution of the quadratic equation (4), then f(x) must be a solution of (4), too. Hence, we have 4 possible cases: Case 1. $ f\left( 1 + \sqrt {2}\right) = 1 + \sqrt {2}$ and $ f\left( 1 - \sqrt {2}\right) = 1 - \sqrt {2}$. Case 2. $ f\left( 1 + \sqrt {2}\right) = 1 - \sqrt {2}$ and $ f\left( 1 - \sqrt {2}\right) = 1 + \sqrt {2}$. Case 3. $ f\left( 1 + \sqrt {2}\right) = 1 + \sqrt {2}$ and $ f\left( 1 - \sqrt {2}\right) = 1 + \sqrt {2}$. Case 4. $ f\left( 1 + \sqrt {2}\right) = 1 - \sqrt {2}$ and $ f\left( 1 - \sqrt {2}\right) = 1 - \sqrt {2}$. The latter two cases are impossible because of the injectivity; hence, one of the Cases 1 and 2 holds. Wherever Case 1 holds or Case 2 is unimportant; in any case, we have $ f\left( 1 + \sqrt {2}\right) + f\left( 1 - \sqrt {2}\right) = 2$ and $ \left| f\left( 1 + \sqrt {2}\right) - f\left( 1 - \sqrt {2}\right) \right| = 2\sqrt {2}$. Hence, applying (1) to $ x = 1 + \sqrt {2}$ and $ y = 1 - \sqrt {2}$, we find $ \left| f\left( \frac{2}{2\sqrt{2}}\right) \right| =\frac{2}{2\sqrt{2}}$, i. e. $ \left| f\left( \frac{1}{\sqrt{2}}\right) \right| =\frac{1}{\sqrt{2}}$, so that, using (2), we have $ \left| f\left( \sqrt {2}\right) \right| = \sqrt {2}$. Therefore, by the multiplicativity, $ f\left( 2\right) =f\left( \sqrt{2}\right) \cdot f\left( \sqrt{2}\right) =\left( f\left( \sqrt{2}\right) \right) ^{2}=\left| f\left( \sqrt{2}\right) \right| ^{2}=2$. Then, we may proceed as you did and show that f(x) = x for any rational x. But how do you get the positivity? Darij

Well, I think that we can derive that from the fact that $ f(xy)=f(x)f(y) $ for any x and y. Just put x=y.

harazi wrote: Well, I think that we can derive that from the fact that $ f(xy) = f(x)f(y)$ for any x and y. Just put x=y. LOL... I must have been really blind not to find this myself! Darij

It's a problem of the france correspondance 2002-2003 See: http://www.animath.fr/tutorat/dossier_02036.html

Valiowk wrote: Find all functions $ f: \mathbb{R} \to \mathbb{R}$ satisfying \[ f\left(\frac {x + y}{x - y}\right) = \frac {f\left(x\right) + f\left(y\right)}{f\left(x\right) - f\left(y\right)} \]for all $ x \neq y$. Pretty neat. Solution. As usual, let \(P(x,y)\) assert the above equation. Then, we prove some claims. Claim 1. \(f\) is injective. Proof. This holds because if \(f(x)=f(y)\) and \(x\neq y\), then the left hand side is defined while the right hand side is not, a contradiction. Claim 2. \(f\) is an odd function. Proof. \(P(x,-x)\) gives us that \[f(0)=\frac{f(x)+f(-x)}{f(x)-f(-x)}\]so \[(f(x)+f(-x))(f(y)-f(-y))=(f(x)-f(-x))(f(y)+f(-y))\]or \[f(x)f(y)-f(x)f(-y)+f(-x)f(y)-f(-x)f(-y)=f(x)f(y)+f(x)f(-y)-f(-x)f(y)-f(-x)f(-y)\]for all \(x,y\) implying that \(\frac{f(x)}{f(-x)}\) is constant, say \(c\). Plugging this back into \(P(x,-y)\) gives us that \(c=-1\) and we are done. Claim 3. \(f\) is multiplicative. Proof. Firstly, \(P(x,0)\) yields that \(f(1)=1\). Next, \(P\left(\frac{x+y}{x-y},1\right)\) gives us that \[f\left(\frac{\frac{x+y}{x-y}+1}{\frac{x+y}{x-y}-1}\right)=\frac{f\left(\frac{x+y}{x-y}\right)+1}{f\left(\frac{x+y}{x-y}\right)-1}\]implying that \(f\left(\frac{x}{y}\right)=\frac{f(x)}{f(y)}\) and so we are done. Claim 4. \(f\) is strictly increasing over \(\mathbb{R}^+\). Proof. We need to show that \(a<b\) implies \(f(a)<f(b)\) for all \(x,y\in\mathbb{R}^+\). \(P(a,-b)\) with \(a<b\) and \(a,b>0\) yields that \[0>f\left(\frac{a-b}{a+b}\right)=\frac{f(a)-f(b)}{f(a)+f(b)}\]and since \(f\) is multiplicative, \(x\geq0\) implies \(f(x)\geq0\) for all positive reals \(x\)(since \(f(x^2)=f(x)^2\)), we see that \(f(a)<f(b)\) and this completes the proof of this claim. Coming back to our problem, since \(f\) is multiplicative, let \(f(x)=g(\ln x)\). Then, since \(x\) is a positive real, \(g(x+y)=g(x)g(y)\) for all \(x,y\in\mathbb{R}^+\). Let \(h(x)=\ln g(x)\). Then, we see that \(h\) is an additive function. Also, since \(f\) is strictly increasing over the positive reals, so is \(g\) and so is \(h\). Therefore, \(h\) is monotone and additive, implying that \(h(x)=kx\) where c is a constant. This means that \(g(x)=e^{kx}\) and so \(f(x)=e^{k\ln x}\) implying that \(f(x)=x^k\) for all \(x\). Plugging in the original equation yields that \(k=1\) or, \(f(x)=x\) for all \(x\in\mathbb{R}^+\). Since \(f\) is odd, we can conclude that \(f(x)=x\) for all \(x\in\mathbb{R}\) which indeed fits.

We claim that the only such function is $f \equiv x$, plugging it in functional equation, we can see that it works. The second condition of our problem says that $f$ is injective. Let $P(x, y)$ be the assertion, we now proceed with a bunch of claims. Claim: $f(0)=0.$ Proof: First of all notice that constant functions don't satisfy our functional equation. Now, $$P(x, 0) \rightarrow f(1) = \frac{f(x)+f(0)}{f(x)-f(0)}$$$$\iff f(x)=f(0)\frac{f(1)+1}{f(1)-1}$$so $f(0)$ must be zero since otherwise the function would be constant. $\blacksquare$ Claim: $f(1)=1.$ Proof: $P(x, 0) \rightarrow f(1)=\frac{f(x)+f(0)}{f(x)-f(0)}=1$ $\blacksquare$ Claim: $f(x)+f(-x)=0$ i.e. $f$ is odd. Proof: $P(x, -x) \rightarrow f(x)+f(-x)=0.$ $\blacksquare$ Claim: For any $x \in \mathbb{R}-\{0\}$, $f(x)f(\frac{1}{x})=1$. Proof: $$P(x, -y) \rightarrow f\left(\frac{x-y}{x+y}\right)=\frac{f(x)+f(-y)}{f(x)-f(-y)}=\frac{f(x)-f(y)}{f(x)+f(y)}$$Combining this with $P(x, y)$, we get our desired result. $\blacksquare$ Claim: $f(xy)=f(x)f(y)$ i.e. $f$ is multiplicative. Proof: $$P(xy, 1) \rightarrow f\left(\frac{xy+1}{xy-1}\right) = \frac{f(x)+f(y)}{f(x)-f(y)}$$$$\iff f(xy)=\frac{f\left(\frac{xy+1}{xy-1}\right)+1}{f\left(\frac{xy+1}{xy-1}\right)-1}$$$$P(x, \frac{1}{y}) \rightarrow \frac{f\left(\frac{xy+1}{xy-1}\right)+1}{f\left(\frac{xy+1}{xy-1}\right)-1}=f(x)f(\frac{1}{y})=f(x)f(y)$$So indeed we have $f(xy)=f(x)f(y)$. $\blacksquare$ Claim: $f$ is strictly increasing over $\mathbb{R}^{+}$ Proof: First of all, notice that $f(x^2)=f(x)^2 \geq 0 \forall x \in \mathbb{R}$ which means $f >0 \forall x \in \mathbb{R}^{+}$. Now suppose we have $a$ and $b$ such that $a>b>0$, $$f\left(\frac{x+y}{x-y}\right)=\frac{f(x)+f(y)}{f(x)-f(y)} >0$$which means $f(x)>f(y)$. $\blacksquare$ Now let $g(x)=\ln(f(a^x))$, notice that $g(x+y)=g(x)+g(y)$. With all the conditions we have, $g(x)=cx$ and this means $f(x)=x^c$ and plugging it in our original equation, we get $c=1$ and $f \equiv x$. $\blacksquare$

Valiowk wrote: Find all functions $ f: \mathbb{R} \to \mathbb{R}$ satisfying \[ f\left(\frac {x + y}{x - y}\right) = \frac {f\left(x\right) + f\left(y\right)}{f\left(x\right) - f\left(y\right)} \]for all $ x \neq y$.

https://artofproblemsolving.com/community/c6h3020980p27158341

The answer is $f(x) = \boxed{x}$, which works. Moreover, denote the given assertion as $P(x,y)$. We will solve this problem through a series of claims: Claim 1: $f(0)=0$ and $f(1)=1$ Proof: Suppose that $f(1) \neq 1$. Notice that $P(x,0)$ yields \[f(x) = \frac{f(0)f(1)+f(0)}{f(1)-1},\] which means $f$ is constant, a contradiction. Hence, $f(1)=1$ and therefore, $f(0)=0.$ $\square$ Claim 2: $f$ is odd Proof: Plug in $P(x,-x)$ to get \[0 = f(0) = \frac{f(x)+f(-x)}{f(x)-f(-x)},\] which simplifies to $f(x) = -f(-x)$, as desired. $\square$ Claim 3: $f$ is multiplicative Proof: Notice that \[\frac{f(x)+f(y)}{f(x)-f(y)} = f \left(\frac{x+y}{x-y} \right) = f \left(\frac{x/y+1}{x/y-1} \right) = \frac{f(x/y)+1}{f(x/y)-1}. \] Clearing denominators and simplifying, we get \[f\left(\frac{x}{y} \right) f(y) = f(x). \ \square\] Claim 4: $f$ is strictly increasing Proof: Since $f$ is odd, we can WLOG only consider when $x \in \mathbb{R}^+$. Then, note that \[f(x) = f(t^2) = f(t)^2 >0,\] for some $t \in \mathbb{R}^+$; this means $f$ is positive over the given domain. If $x>y>0$, then \[\frac{f(x)+f(y)}{f(x)-f(y)} = f\left(\frac{x+y}{x-y}\right) > 0,\] implying that $f(x)>f(y)$. $\square$ Claim 5: $f(x) = x$ for all $x \in \mathbb{Z}$. Proof: We will prove this using induction on $x>0$. For the first base case, observe that $f(1)=1$. As for the other case, suppose that $f(2) = a$. Consequently, \begin{align*} f(3) &= f\left(\frac{2+1}{2-1}\right) = \frac{a+1}{a-1} \\ f(4) &= f(2)^2 = a^2 \\ f\left(\frac{5}{3} \right) &= f\left(\frac{4+1}{4-1}\right)= \frac{a^2+1}{a^2-1} \\ f(5) &= f\left(\frac{3+2}{3-2}\right) = \frac{a^2+1}{-a^2+2a+1}. \end{align*} Then, \[\frac{a+1}{a-1} \cdot \frac{a^2+1}{a^2-1} = f(3) f\left(\frac{5}{3} \right) = f(5) = \frac{a^2+1}{-a^2+2a+1}.\] Solving this yields $a=0$ or $a=2$, but the former is impossible due to $f$ being increasing. Hence, $f(2)=2$, completing the base case. For the inductive step, consider $P(x,1)$: \[f\left(\frac{x+1}{x-1}\right) = \frac{f(x+1)}{f(x-1)} = \frac{f(x)+1}{f(x)-1}.\] It is easy to see that if $f(k-1) =k-1$ and $f(k)=k$ by the inductive hypothesis, then $f(k+1)=k+1$, finishing the induction. $\square$ Because $f$ is multiplicative, $f$ is also equivalent to the identity function over rationals as well; furthermore, $f$ is strictly increasing, which expands the identity function to all real numbers.

Note that $$f(1)=\frac{f(x)+f(0)}{f(x)-f(0)}\Rightarrow f(x)(1-f(1))=f(0)(1+f(1)) \qquad (1)$$since it is clear that $f$ can't be constant, it follows from $(1)$ that $f(1)=1$. We also get $f(0)=0$ from $(1)$. Now putting $y=-x$ in the main equation, we find $$\frac{f(x)+f(-x)}{f(x)-f(-x)}=f(0)=0 \Rightarrow f(-x)=-f(x).$$So $f$ is an odd function. Now, notice that $$\frac{f(x)+1}{f(x)-1}=f\left(\frac{x+1}{x-1}\right)=f\left(\frac{xy+y}{xy-y}\right)=\frac{f(xy)+f(y)}{f(xy)-f(y)} \Rightarrow f(xy)=f(x)f(y).$$So $f$ is multiplicative. Next, we square things using the multiplicativity: $$f\left(\frac{x+y}{x-y}\right)^2=\left(\frac{f(x)+f(y)}{f(x)-f(y)}\right)^2=\frac{f(x)^2+2f(x)f(y)+f(y)^2}{f(x)^2-2f(x)f(y)+f(y^2)}=\frac{f(x^2)+f(y^2)+f(2xy)}{f(x^2)+f(y^2)-f(2xy)} \qquad (2)$$and $$f\left(\left(\frac{x+y}{x-y}\right)^2\right)=\frac{f\left((x+y)^2\right)}{f\left((x-y)^2\right)}=\frac{f(x^2+y^2+2xy)}{f(x^2+y^2-2xy)}=\frac{f(x^2+y^2)+f(2xy)}{f(x^2+y^2)-f(2xy)}\qquad (3)$$Comparing $(2)$ and $(3)$, we see that $f(x^2+y^2)=f(x^2)+f(y^2)$ for all $x\neq y$. But this holds for $x=y$ too for if we put $y=-x$ in $(2)$ then we get $f(2x^2)=2f(x^2)$. So $f$ is additive over $\mathbb R_{>0}$, but since $f$ is odd, it must be additive over $\mathbb R$. Since $f$ is additive, multiplicative and non constant, $f$ must be the identity function. Clearly, $f(x)=x$ for all $x$ satisfies our equation. So this is the only solution.

We claim $f(x)=x$ is the only solution, which clearly works. First $P(x,0)$ implies $f(1)=1$ and $f(0)=0$ or $\frac{f(x)}{f(0)}=\frac{f(1)+1}{f(1)-1}$ contradicting injectivity. Now $P(x,-x)$ implies $f$ is odd. Taking $P(x,kx)$ implies $\frac{f(kx)}{f(x)}=\frac{f\left(\frac{1+k}{1-k}\right)-1}{f\left(\frac{1+k}{1-k}\right)+1}$ for all $x\ne 0$. In particular $f(k^2)=\frac{f(k^2)}{f(k)}\cdot\frac{f(k)}{f(1)}=\left(\frac{f\left(\frac{1+k}{1-k}\right)-1}{f\left(\frac{1+k}{1-k}\right)+1}\right)^2\ge 0$, so $x>0$ implies $f(x)>0$. Next, $P(1\pm\sqrt2,1)$ gives $f(1\pm\sqrt2)=1\pm\sqrt2$. Then $P(\sqrt2+1,\sqrt2-1)$ gives $f(\sqrt2)=\pm\sqrt2$, so $f(2)=2$. We now show by strong induction that $f(n)=n$ for positive integers $n$, and thus by oddness all integers. It is true for $n=1,2$. If it is true up to $n-1$ then if $n$ is even write $f(n)=\frac{f(n)}{f\left(\frac n2\right)}\cdot\frac{f\left(\frac n2\right)}{f(1)}=2\cdot\frac n2=n$. If $n$ is odd then $P\left(\frac{n+1}2,\frac{n-1}2\right)$ gives $f(n)=n$. Next if $r=\frac ab$ is rational for integers $a,b$ then $P(a+b,a-b)$ gives $f(r)=r$. Finally, if $x>y\ge 0$ then $P(x,y)$ implies $f(x)>f(y)$. By oddness, $f$ is increasing. Since it is the identity on rationals, it is the identity on reals.