Put $ w = x = y = z = t$ and observe that $ f(t^{2}) = f(t)^{2}$. Set $ k(t) = \frac {f(t)}{t}$ and consider $ pq = s ^{2}$. You get $ \frac {f(s^{2})}{s^{2}} = k(s^{2}) = \frac {f(p^{2}) + f(q^{2})}{p^{2} + q^{2}}$ or $ \frac {f(s)}{s} = \frac { f(p) + f(\frac {s^{2}}{p}) }{p + \frac {s^{2}}{p}}$ independent of p. Now you can let $ p = 1$ and solve the quadratic equation resulting. The solutions are of form $ \frac {s + s^{ - 1} + / - \sqrt {s^{2} + s^{ - 2} + 2 - 4f(1)}}{2}$, and putting $ s = 1$ gives $ f(1)(f(1) - 1) = 0$ so $ f(1) = 1$ and you can get $ f(s)$ taking one of the values $ s$ or $ s^{ - 1}$ at each value of $ s$. Now partition the positive reals (other than 1) into those for which $ f$ gives the identity (typical element p), and those for which $ f$ gives the inverse (typical element $ q$). The original equation contradicts the statement $ f(p^{2}q^{2}) \in ({{p^{2}q^{2},p^{ - 2}q^{ - 2}}})$, so one of those partitions must be empty. Now you can go ahead and verify that either identity (everywhere) or inverse (everywhere) are satisfactory.

the.sceth wrote: You get $ \frac {f(s^{2})}{s^{2}} = k(s^{2}) = \frac {f(p^{2}) + f(q^{2})}{p^{2} + q^{2}}$ or \[ \frac {f(s)}{s} = \frac { f(p) + f(\frac {s^{2}}{p}) }{p + \frac {s^{2}}{p}} \] independent of p. Sorry, how do you get from the first equation to the second? Did you replace $ p^2,q^2,s^2$ with $ p,q,s$?

Here is my solution . Take $ x = y = z = w$ then we have \[ f(x^2) = f(x)^2 \Rightarrow f(1) = 1 \] Take $ x\to \sqrt {x},y\to \sqrt {y},z\to sqrt{z},w\to \sqrt {w}$ we have : \[ (y + z)(f(x) + f(w)) = (x + w)(f(z) + f(y)) \] Take $ x = w\sqrt {yz}$ then we have : \[ (y + z)f(\sqrt {yz}) = \sqrt {yz}(f(x) + f(y)) \] Take $ z\to 1$ then : \[ \frac {f(x)}{x} = \frac {f(x)^2)}{x^2 + 1} \Leftrightarrow (f(x) - x)(x.f(x) - 1) = 0 \Leftrightarrow f(x) = x,f(x) = \frac {1}{x} \] Suppose exist $ a,b$ is different from 1 such that $ f(a) = a,f(b) = \frac {1}{b}$ Take $ w = 1,x = yz$ then $ (y + z)(f(yz) + 1) = (yz + 1)(f(y) + f(z))$ Take $ y\to a,z\to b$ then \[ (a + b)(f(ab) + 1) = (ab + 1)(a + \frac {1}{b}) \] If $ f(ab) = ab$ then $ b = \frac {1}{b} \Leftrightarrow b = 1$ If $ ab = \frac {1}{ab}$ then : \[ (a + b) = a(ab + 1) = b(ab + 1) \] So $ a = b$ Therefore $ a = b = 1$ (contradiction ) There are only two function satisfy condition : 1) $ f(x)\equiv x$ 2)$ f(x)\equiv \frac {1}{x}$

My solution (at home) is very similar to TTsphn's one...but quite longer and messier By the way nice solution! Daniel

orl wrote: Find all functions $ f: (0, \infty) \mapsto (0, \infty)$ (so $ f$ is a function from the positive real numbers) such that \[ \frac {\left( f(w) \right)^2 + \left( f(x) \right)^2}{f(y^2) + f(z^2) } = \frac {w^2 + x^2}{y^2 + z^2} \] for all positive real numbes $ w,x,y,z,$ satisfying $ wx = yz.$ Author: Hojoo Lee, South Korea hi for $ x=y=z=w f(x^2)=f(x)^2$ so $ f(1)=1 (f(x)>0)$ supposing that $ x=w$so $ x^2=yz$ that mean: $ f(x^2)(y^2+z^2) =x^2 (f(y^2)+f(z^2))$ $ f(yz) (y^2+z^2 )= yz (f(y^2) +f(z^2))$ for $ y=1$ we have : $ f(z)(z^2+1)=z(f(z^2)+1)$ wich include $ f(z)(z^2+1)=z(f(z)^2+1)$ ==> $ zf(z) (z-f(z))+(f(z)-z) =0$ $ (z-f(z) )(zf(z)-1)=0$==>$ f(z)=z$ or $ f(z)=(1/z)$

Note that you are not done there: there could be a horribly convoluted and nasty function which maps $ x$ to $ x$ for some real $ x$ and $ x$ to $ \frac {1}{x}$ for others. There remains to prove that no such function satisfying the conditions of the problem exists (as TTsphn did, for example).

Lepuslapis wrote: Note that you are not done there: there could be a horribly convoluted and nasty function which maps $ x$ to $ x$ for some real $ x$ and $ x$ to $ \frac {1}{x}$ for others. There remains to prove that no such function satisfying the conditions of the problem exists (as TTsphn did, for example). yes you are right we have to find contradiction for the case of f determinated us $ f$ : $ f(x)=x$ for $ x \in [0.a[$ $ f(x)=\frac{1} {x}$ for $ x \in [a.+00[$

not_trig wrote: the.sceth wrote: You get $ \frac {f(s^{2})}{s^{2}} = k(s^{2}) = \frac {f(p^{2}) + f(q^{2})}{p^{2} + q^{2}}$ or \[ \frac {f(s)}{s} = \frac { f(p) + f(\frac {s^{2}}{p}) }{p + \frac {s^{2}}{p}} \] independent of p. Sorry, how do you get from the first equation to the second? Did you replace $ p^2,q^2,s^2$ with $ p,q,s$? Exactly. Also, bear in mind that the idea of the partition was equivalent to TTsphn´s method of eliminating the posibility of $ f(t)$ being somewhere $ f(t) = t^{ - 1} \neq 1$ and somewhere $ f(t)=t \neq 1$. (I.e. as Lepuslapis pointed out.) It´s just with computation omitted.

as posted before $ f(x^2)=f^2(x)$.. we can substitute this at the hypothesis and get $ \dfrac{f(a)+f(b)}{f(c)+f(d)}=\dfrac{a+b}{c+d}$ with $ ab=cd$... so, $ \dfrac{f(x^3)+f(x)}{2f(x^2)}=\dfrac{x^2+1}{2x}$, and it follows that $ f(x^3)=f(x)\left(\dfrac{f(x)(x^2+1)}{x}-1\right)$. we also have that $ \dfrac{f(x^4)+f(x)}{f(x^2)+f(x^3)}=\dfrac{x^2-x+1}{x}$ if we let $ f(x)=a$ we have that $ f(x^4)=(\dfrac{x^2-x+1}{x})(a^2+\dfrac{a^2(x^2+1)}{x}-a)-a=a\cdot \left(\dfrac{x^2-x+1}{x}\cdot\left(\dfrac{a(x^2+x+1)}{x}-1\right)-1\right)$ but $ f(x^4)=f^2(x^2)=f^4(x)=a^4$, so it follows that $ a^3=\dfrac{x^2-x+1}{x}\cdot\left(\dfrac{a(x^2+x+1)}{x}-1\right)-1$ this cubic equation has a negative solution and $ a=x$ and $ a=\dfrac{1}{x}$... and this can be finished as any of the solutions above...

By putting $ w=x=y=z$ we have $ f(x^2)=f(x)^2$ for all $ x$. For $ x=1$ we have $ f(1)=1$. So $ f(w)^2+f(z)^2 = f(w^2)+f(z^2)$, and then $ \frac{f(x^2)+f(y^2)}{x^2+y^2}$ depends only on $ xy$, and so does $ \frac{f(x)+f(y)}{x+y}$. So we can say that $ \frac{f(x)+f(y)}{x+y} = \frac{f(xy)+f(1)}{xy+1} = \frac{f(xy)+1}{xy+1}$. By putting $ x=y$ and replacing $ f(x^2)$ by $ f(x)^2$ we get the quadratic equation in $ f(x)$ $ x\cdot f(x)^2 - (x^2+1)\cdot f(x) + x=0$ that has solutions $ f(x)=x$, $ f(x)=\frac{1}{x}$. It remains to control that there can't be $ a,b\neq 1$ so that $ f(a)=a$ and $ f(b)=\frac1b$, as TTsphn did.

Well i did the same for $ f(x^2) = (f(x))^2$. Then $ f(1) = 1$ and so $ \displaystyle f(x) + f\left(\frac {1}{x}\right) = x + \frac 1 x$ (*). Now plug $ w = a^2, x = 1, y = a,z = a$ and we get $ \displaystyle \frac 1{f(x)} + f(x) = x + \frac 1x$(**) and since $ g(x) = x + \frac 1x$ is a strictly increasing function in $ (1,\infty)$ we obtain that $ f(x) = x$ or $ \displaystyle f(x) = \frac 1x$. From (*) and (**) we get that $ \displaystyle \frac 1{f(x)} = f(\frac 1x)$(***). Now to conclude use (***) suposing $ f(a) = a$, $ f(b) = \frac 1b$ and using the fact that $ f(x^{2^k}) = (f(x))^{2^k}$ we can assume that $ a$ and $ b$ are small enough or large enough to find that $ f(ab) \not = ab$ and $ \displaystyle f(ab) \not = \frac 1{ab}$ (pluging $ w = 1, x = ab, y = a, z = b$) which is a contradiction (in fact taking a large $ b$ and and choosing $ a$ such that $ ab<1$ one gets that $ \displaystyle f(ab)= \frac {ab + a +\frac 1b - b}{a+b}<0$).

orl wrote: Find all functions $ f: (0, \infty) \mapsto (0, \infty)$ (so $ f$ is a function from the positive real numbers) such that \[ \frac {( f(w) )^2 + ( f(x) )^2}{f(y^2) + f(z^2) } = \frac {w^2 + x^2}{y^2 + z^2} \] for all positive real numbes $ w,x,y,z,$ satisfying $ wx = yz.$ Author: Hojoo Lee, South Korea

This problem seems not very hard

As before, show that $ f(x)^2 = f(x^2)$. Also similarly, substitute $ a^2, b^2, 1, ab$ to show that $ P(a, b) : f(a^2) + f(b^2) = f(ab) \left( \frac{a}{b} + \frac{b}{a} \right)$. Now $ P(a, b) - P(\sqrt{a}, \sqrt{b})^2 : f(a^2) f(b^2) = f(ab)^2$. The same quadratic-solving step (letting $ ab = 1$) shows that $ f(a) = a \text{ or } \frac{1}{a} \forall a$, but now note that if $ f(a) = a, f(b) = \frac{1}{b}$ then $ \frac{a}{b} = ab \text{ or } \frac{1}{ab}$ which is clearly only possible if $ b = 1$ or $ a = 1$, respectively.

gauss_chen wrote: This problem seems not very hard Of course! Because it was chosen to be at the easy level !

Too easy $ x=y=z=w \implies f(x^2)=(f(x))^2 \implies f(1)=1$ now set $ w=1$, $ y=z=\sqrt{x}$..... and the rest is like TTsphn.

TTsphn wrote: .... If $ ab = \frac {1}{ab}$ then : \[ (a + b) = a(ab + 1) = b(ab + 1) \] So $ a = b$ Therefore $ a = b = 1$ (contradiction ) There are only two function satisfy condition : 1) $ f(x)\equiv x$ 2)$ f(x)\equiv \frac {1}{x}$ It must be :" If $ f(ab) = \frac {1}{ab}$ then : .... "

not to bump up on an old topic but i did this problem more easily than any post ive seen is this correct? f(1)=1 2f(0)/((f(0)+f(1))=1 => f(0)=0 f(n)^2/f(1)=n^2 f(n)=n,-n and trying both we see that only n works so QED?

Note zero is not in the domain nor the codomain of the function.

Pretty easy for p4. First of all, letting all terms equal, find that $f(x)^2$=$f(x^2)$. Then, observe that $P(x,x,1,x^2)$ yields that $f(x)(x^2+1)=x(f(x)^2+1)$ for all positive real numbers $x$.(We get $f(1)=1$) Note that,if the function $g(x)=(x^2+1)/x$ has 2 equal values for 2 inputs $x$ and $y$, then either $x=y$ or $y=1/x$. So, $f(x)$ is either $x$ or $1/x$ for all $x$. Taking $a,b$ such that neither of them is $1$, let $P(a,b,1,ab)$. Messing with those 2 nasty cases gives us a contradiction. Therefore, $f(x)=x$ for all $x$ or $f(x)=1/x$ for all $x$.

Taking all the variables to be $1$, $f(1)=1$. Afterwards, set the variables to be equal, so $f(t)^2=f(t^2)$. Letting $(w, x, y, z)=(t, 1, \sqrt t, \sqrt t)$ gives \[\frac{f(t)^2+1}{2f(t)}=\frac{t^2+1}{2t}.\]It's easy to see that $f(t)=t$ and $f(t)=\tfrac1t$ are solutions -- since this is a quadratic in $f(t)$, there are at most $2$ solutions. Now suppose that we have $f(a)=a$ and $f(b)=\tfrac1b$, for $a$, $b\neq 1$. Take $(w, x, y, z)$ as $(\sqrt a, \sqrt b, 1, \sqrt{ab})$. So \[\frac{f(\sqrt{a})^2+f(\sqrt b)^2}{f(\sqrt{ab})^2+1}=\frac{a+b}{ab+1}.\]The LHS is also $\tfrac{f(a)+f(b)}{f(ab)+1}$, hence \[\frac{a+\tfrac1b}{f(ab)+1}=\frac{a+b}{ab+1}.\]There are two cases for the value of $f(ab)$ -- it's either $ab$ or $\tfrac1{ab}$. $f(ab)=ab$: we get $b^2=1$ implying $b=1$. Contradiction. $f(ab)=\tfrac{1}{ab}$: we get $a^2=1$ so $a=1$. Contradiction. It's easy to verify that $f(t)=t$ and $f(t)=\tfrac1t$ both work. So we are done. $\square$

We claim that $f(x) = x$ or $f(x) = \frac{1}{x}$, which both clearly work. Now, substituting $w = x = y = z$ yields \[f(x)^2 = f(x^2)\text{.}\]This implies that $f(1) = 1$. Now substitute $w = 1, x = n, y = z = \sqrt{n}$ for some positive real number $n$. This yields \begin{align*} \frac{f(1)^2 + f(n)^2}{2f(n)} &= \frac{1 + n^2}{2n} \\ n(1 + f(n)^2) &= f(n)(1 + n^2) \\ f(n)^2 + \left(n + \frac{1}{n}\right) f(n) + 1 &= 0 \\ (f(n) - n)\left(f(n) - \frac{1}{n}\right) &= 0\text{.} \end{align*}Therefore, for each $n$, $f(n) = n$ or $\frac{1}{n}$. Now suppose there exist $a \neq b$, none of which are equal to $1$, such that $f(a) = a$ and $f(b) = \frac{1}{b}$. Now \begin{align*} \frac{f(1)^2 + f(ab)^2}{f(a^2) + f(b^2)} &= \frac{1 + (ab)^2}{a^2 + b^2} \\ \frac{1 + f(ab)^2}{f(a)^2 + f(b)^2} &= \frac{1 + (ab)^2}{a^2 + b^2} \\ 1 + f(ab)^2 &= \frac{(1 + (ab)^2)(a^2 + \frac{1}{b^2})}{a^2 + b^2} \\ (1 + f(ab)^2)(a^2 + b^2) &= 2a^2 + \frac{1}{b^2} + a^4 b^2 \\ a^2 + b^2 + f(ab)^2 a^2 + f(ab)^2 b^2 &= 2a^2 + \frac{1}{b^2} + a^4 b^2\text{.} \end{align*}If $f(ab) = ab$, then we have \begin{align*} a^2 + b^2 + a^4 b^2 + a^2 b^4 &= 2a^2 + \frac{1}{b^2} + a^4 b^2 \\ b^2 + a^2 b^4 &= a^2 + \frac{1}{b^2} \\ a^2 (b^4 - 1) + \frac{1}{b^2} (b^4 - 1) &= 0 \\ \left(a^2 + \left(\frac{1}{b}\right)^{\!2}\right)(b^4 - 1) &= 0 \end{align*}which is clearly impossible. Similarly, if $f(ab) = \frac{1}{ab}$, then we have \begin{align*} a^2 + b^2 + \frac{1}{b^2} + \frac{1}{a^2} &= 2a^2 + \frac{1}{b^2} + a^4 b^2 \\ a^2 + a^4 b^2 &= b^2 + \frac{1}{a^2} \\ b^2 (a^4 - 1) + \frac{1}{a^2} (a^4 - 1) &= 0 \\ \left(b^2 + \left(\frac{1}{a}\right)^{\!2}\right)(a^4 - 1) &= 0 \end{align*}which is also impossible.

Plug in $y=w, z=x$ $f(x)^2+f(y)^2=f(x^2)+f(y^2)$ When $x=y$, get that $f(x)^2=f(x^2)$ So plugging in $x=1$, get that $f(1)=1$ If $w=1$, then $x=yz$ $\frac{1+f(yz)^2}{f(y^2)+f(z^2)}=\frac{1+y^2z^2}{y^2+z^2}$ If $y=z=\sqrt{x}$, then $\frac{1+f(x)^2}{2f(x)}=\frac{1+x^2}{2x}$ $2x+2xf(x)^2=2f(x)+2x^2f(x)$ $2xf(x)^2+f(x)(-2x^2-2)+2x=0$ $f(x)=x$ or $\frac{1}{x}$ by the quadratic formula

The answers are $f(x) = x$ and $f(x) = \tfrac{1}{x}$, both of which aren't too hard to verify. We now show these are the only answers. Let $P(w, x, y, z)$ denote the given assertion. From $P(1, 1, 1, 1)$, we have $f(1) = 1$. Then, from $P(1, x, \sqrt{x}, \sqrt{x})$, we have $\tfrac{1 + f(x)^2}{2f(x)} = \tfrac{1 + x^2}{2x}.$ This rearranges to $$xf(x)^2 - f(x) - x^2f(x) + x = 0 \iff (xf(x) - 1)(f(x) - x) = 0.$$Thus, for all $x$, either $f(x) = \tfrac{1}{x}$ or $f(x) = x$. Now, suppose for the sake of contradiction there exist $a, b > 0$ and not equal to $1$ such that $f(a) = \tfrac{1}{a}$ and $f(b) = b$. Then from $P(a, b, 1, ab)$, we see that $$\frac{\tfrac{1}{a^2} + b^2}{1 + f(a^2b^2)} = \frac{a^2 + b^2}{1 + a^2b^2}.$$Taking cases on whether $f(a^2b^2) = \tfrac{1}{a^2b^2}$ or $f(a^2b^2) = a^2b^2,$ we see that: If $f(a^2b^2) = \tfrac{1}{a^2b^2}$, then $\tfrac{1}{a^2} + b^2 + b^2 + a^2b^2 = a^2 + b^2 + \tfrac{1}{a^2} + \tfrac{1}{b^2}$, or $(a^2b^2 + 1)(b^4 - 1) = 0$, which implies that $b = 1$, contradiction If $f(a^2b^2) = a^2b^2$, then $\tfrac{1}{a^2} + b^2 + b^2 + a^2b^4 = a^2+ b^2 + a^2b^4 + a^4b^2$, or $(a^2b^2 + 1)(a^4 - 1) = 0$, so $a = 1$, contradiction So, either $f(x) =x$ for all $x$, or $f(x) = \tfrac{1}{x}$ for all $x$, as desired.

I claim that the only solutions are $f(x)=x \forall x\in \mathbb{R}_{>0}$ and $f(x)=\frac{1}{x} \forall x\in \mathbb{R}_{>0}$. (1) First, if we plug in $(c,c,c,c)$, we get that \[\frac{2f(c)^2}{2f(c^2)}=1 \implies f(c)^2=f(c^2),\]since $f$ is over the positive reals. Additionally, we can also plug in $c=1$ to get that \[f(1)^2=f(1) \implies f(1)=1,\]again since $f$ is over the positive reals. (2) Now, if we plug in $(\sqrt{c},\sqrt{c},1,c)$, we get that \[\frac{f(\sqrt{c})^2+f(\sqrt{c})^2}{f(1)+f(c^2)}=\frac{c+c}{1+c^2} \implies \frac{2f(c)^2}{f(1)+f(c)^2}=\frac{2c}{1+c^2},\]since we found by (1) that $f(x)^2=f(x^2)$. Since $c$ is fixed, letting $f(c)=k$, we can solve the equation \[\frac{k}{1+k^2}=\frac{c}{1+c^2},\]\[\iff k(1+c^2)=c(1+k^2),\]\[\iff ck^2-(1+c^2)k+c=0,\]\[\iff (ck-1)(k-c)=0.\]This means that $k=c$ or $\frac{1}{c}$, which implies that $\forall x\in \mathbb{R}_{>0}$, $f(x)$ is either $x$ or $\frac{1}{x}$. Now for the pointwise trap! I claim that either $f(x)=x \forall x\in \mathbb{R}_{>0}$ or $f(x)=\frac{1}{x} \forall x\in \mathbb{R}_{>0}$. FTSOC, assume that $\exists a$, $b\neq 1$ such that $f(a)=a$ and $f(b)=\frac{1}{b}$. Plugging in $(\sqrt{ab},\sqrt{ab},a,b)$ then gives us that \[\frac{f(ab)+f(ab)}{f(a)^2+f(b)^2}=\frac{ab+ab}{a^2+b^2} \implies \frac{f(ab)}{a^2+\frac{1}{b^2}}=\frac{ab}{a^2+b^2}, \]\[\iff f(ab)=\left(\frac{ab}{a^2+b^2}\right)\left(\frac{a^2b^2+1}{b^2}\right).\]By (2), we know that $\forall x\in \mathbb{R}_{>0}$, $f(x)$ is either $x$ or $\frac{1}{x}$, which means that $f(ab)=ab$ or $\frac{1}{ab}$. If $f(ab)=ab$, then we have that \[\left(\frac{ab}{a^2+b^2}\right)\left(\frac{a^2b^2+1}{b^2}\right)=ab,\]\[\iff \frac{a^2b^2+1}{b^2(a^2+b^2)}=1,\]\[\iff a^2b^2+1=b^2(a^2+b^2) \implies a^2b^2+1=a^2b^2+b^4,\]\[\iff b^4=1,\]\[\implies b=1,\]since $f$ is over the positive reals. However, this is a contradiction to our requirement that $a$, $b\neq 1$. Therefore, we get that $f(ab)=\frac{1}{ab}$. Now, if $f(ab)=\frac{1}{ab}$, we get that \[\left(\frac{ab}{a^2+b^2}\right)\left(\frac{a^2b^2+1}{b^2}\right)=\frac{1}{ab},\]\[\iff ab(ab)(a^2b^2+1)=b^2(a^2+b^2),\]\[\iff a^4b^4+a^2b^2=a^2b^2+b^4,\]\[\iff a^4b^4=b^4,\]\[\implies a^4=1,\]\[\implies a=1,\]since $f$ is over the positive reals. This is again a contradiction to our requirement that $a$, $b\neq 1$, meaning that there cannot exist $a$, $b\neq 1$ such that $f(a)=a$ and $f(b)=\frac{1}{b}$. Therefore our only solutions are $f(x)=x \forall x\in \mathbb{R}_{>0}$ and $f(x)=\frac{1}{x} \forall x\in \mathbb{R}_{>0}$, which can both be checked to work through plugging and chugging, finishing the problem.

Solved with blueberryfaygo_55. We claim either $\boxed{f(x)\equiv x}$ or $\boxed{f(x)\equiv\frac{1}{x}}$. It is easy to check that these work. Let $P(w,x,y,z)$ be the given assertion. Then $P(1,1,1,1)$ gives $f(1)=1$. $P(x,x,x,x)$ gives $f(x)^2=f(x^2)$. It follows that \[ \frac{f(w^2)+f(x^2)}{f(y^2)+f(z^2)}=\frac{w^2+x^2}{y^2+z^2} \]whenever $wx=yz$. Letting $(a,b,c,d):=(w^2,x^2,y^2,z^2)$, we have \[ \frac{f(a)+f(b)}{f(c)+f(d)}=\frac{a+b}{c+d}\tag{*} \]whenever $ab=cd$. Plugging in $(a,b,c,d)=(x,x,x^2,1)$ gives \[ \frac{2f(x)}{f(x^2)+1}=\frac{2x}{x^2+1} \]so \[ (x^2+1)f(x)=xf(x)^2+x. \]Solving this quadratic yields $f(x)\in\left\{x,\frac{1}{x}\right\}$. Assume for the sake of contradiction there exists $m,n\in\mathbb{Z}^+\setminus\{1\}$ such that $f(m)=m$ and $f(n)=\frac{1}{n}$. Plugging $(a,b,c,d)=(m,n,mn,1)$ into $(*)$ gives \[ \frac{m+\frac{1}{n}}{f(mn)+1}=\frac{m+n}{mn+1}. \]Note that $f(mn)\in\left\{mn,\frac{1}{mn}\right\}$. If $f(mn)=mn$, then \[ \frac{m+\frac{1}{n}}{mn+1}=\frac{m+n}{mn+1} \]so $n=1$, a contradiction. If $f(mn)=\frac{1}{mn}$, then \[ mn\cdot\frac{m+\frac{1}{n}}{1+mn}=\frac{m+n}{mn+1} \]so $m=1$, a contradiction. The conclusion follows. $\square$

Taking $w=x=y=z=1$ gives us that $f(1) = f(1)^{2},$ so $f(1) = 1.$ Also, $w = x = y = z = n$ for some real number $n$ gives us that $f(x^{2}) = f(x)^{2}.$ Now, setting $w = n^{2}, x = 1, y = n, z = n,$ gives us that $$f(n) = n, \frac{1}{n}$$after solving. Now, examining all of the cases, we find that a pointwise trap cannot occur, so we are done.

Our answers are $f(x)=x$ and $f(x)=\frac1x$, and it is easy to check that these satisfy our equation. Plugging in $w=x=y=z$, we get $f(x)^2=f(x^2)$. Therefore, our equation becomes $\frac{f(w^2)+f(x^2)}{f(y^2)+f(z^2)}=\frac{w^2+x^2}{y^2+z^2}$, and since this is over the positive reals, we can replace $w^2$, $x^2$, $y^2$, and $z^2$ with $w$, $x$, $y$, and $z$ respectively, giving us \[\frac{f(w)+f(x)}{f(y)+f(z)}=\frac{w+x}{y+z}.\]Note that this substitution works with the condition $wx=yz$ because previously, this implied $w^2x^2=y^2z^2$, meaning replacing the variables still makes our equation hold true for $wx=yz$. Plugging in $w=x=y=z=1$ into the original equation, we get $\frac{2f(1)^2}{2f(1)}=1$, giving $f(1)=1$. Now, plugging in $1$, $x^2$, $x$, and $x$ for $w$, $x$, $y$, and $z$ respectively, we get \[\frac{f(1)+f(x^2)}{2f(x)}=\frac{1+f(x^2)}{2f(x)}=\frac{1+x^2}{2x}.\]Clearing the denominators and factoring this, we get \[2(f(x)-x)(xf(x)-1)=0,\]which gives $f(x)=x$ and $f(x)=\frac1x$. We will show that we cannot have $f(a)=a$ while $f(b)=\frac{1}{b}$ at the same time. Suppose $a,b\neq 1$, because otherwise we would have $a=\frac1a$ or $b=\frac1b$. Then, plugging in $a$, $b$, $1$, and $ab$ as $w$, $x$, $y$, and $z$, respectively, we get \[\frac{a+\frac1b}{1+f(ab)}=\frac{a+b}{1+ab},\]and by expanding and simplifying, it is easy to check that this equation does not hold for $f(ab)=ab$ or $f(ab)=\frac{1}{ab}$, so we are done.

The answers are $x, \frac 1x$, both of which clearly work. Set $w = x = y = z = 1$, we get $f(1) = 1$. Now set $y = z = 1$, we get $f(w)^2+ f(\frac 1w)^2 = w^2 + \frac{1}{w^2}$, set $w = x = 1$ and we get $f(y^2) + f(\frac{1}{y^2}) = y^2 + \frac{1}{y^2}$. Thus we have $f(k)^2 + f(\frac 1k)^2 = k^2 + \frac{1}{k^2}$, as well as $f(k) + f(\frac 1k) = k + \frac 1k$, squaring the latter equation and subtracting from the first yields $f(k)f(\frac 1k) = 1$. Now using this with $f(k) + f(\frac 1k) = k + \frac 1k$ and solving the resulting quadratic gives $f(k) = k$ OR $f(k) = \frac 1k$ for each individual $k$. It remains to resolve the pointwise trap. We show that if there exist values such that $x \neq 1, f(x) = x$ there exist no values with $x \neq 1, f(x) = \frac 1x$, thus we must have either $f(x) = x$ or $f(x) = \frac 1x$ for all $x$. Take $y = z, y \neq 1$, such that $f(y^2) = y^2$, this gives $f(w)^2 + f(x)^2 = w^2 + x^2$. We show that there exists no value of $x \neq 1$ satisfying $f(x) = \frac 1x$. Assume FTSOC this exists, then if $f(w) = w$, we are forced to have $x^4 = 1$, giving $x =1$, otherwise if $f(w) = \frac 1w$ we have $wx = 1$, contradiction since $wx = yz = y^2 \neq 1$.

Here's a sketch: Set $a=b=c=d=1$, the equation implies $f(1)=1$. Set $c=a$ and $d=b=1$ then we get $f(a^2)=f(a)^2$. Set $b=a$, $c=1$ and $d=a^2$ then after a bit of simplification we get $f(a)\in \{1/a,a\}$. Suppose $f(a)=a$ and $f(b)=1/b$ for some $a,b\neq 1$, setting $c=1$, $d=ab$ will give a contradiction (remember $f(ab)=ab$ or $f(ab)=1/(ab)$). Clearly $f(x)=x$ for all $x$ and $f(x)=1/x$ for all $x$ are solutions to the FE.

P(t,t,t,t) gives f(t^2) = f(t)^2 or f(1) = 1 [I proved it in a lil non straight manner but Ic that this works too] P(sqrt(t),sqrt(t), t , 1) gives:- 2f(t)/f(t)^2 + 1 = 2t/t^2+1 or tf(t)^2 - (t^2 + 1)f(t) + t = 0 Solving it as a quadratic (this is possible because t has a degree of freedom at all points) we get:- FINAL ANSWER: f(t)= t, 1/t for all t in R+ both of which can be easily verified

Let $P(w, x, y, z)$ denote the assertion. Then $P(x, x, x, x) \implies f(x^2)=f(x)^2 \implies f(1)=1.$ Also, the assertion then becomes $$\frac{f(w)+f(x)}{w+x} = \frac{f(y)+f(z)}{y+z}.$$Now, note that $$P(x, x, 1, x^2) \implies \frac{f(x)}{x} = \frac{1+f(x^2)}{1+x^2} = \frac{1+f(x)^2}{1+x^2}.$$Therefore, $$0=xf(x)^2-(1+x^2)f(x)+x \implies \left(f(x)-x \right)\left(f(x)-\frac{1}{x}\right)=0,$$so $f(x)$ is $x$ or $\frac{1}{x}$ for each $x \in (0, \infty).$ If there does not exist a $k$ with $f(k)=k,$ then $f(x)=\frac{1}{x}$ for all $x,$ which is a valid solution. Otherwise, for $k, a, b \neq 1$ in our domain with $ab=k^2$ and $f(k)=k,$ note that $$P(k, k, a, b) \implies f(a)+f(b)=a+b.$$If $f(a)=a,$ then $f(b)=b,$ and vice versa. If $f(a)=\frac{1}{a}, f(b)=\frac{1}{b},$ we have that $\frac{1}{a}+\frac{1}{b}=a+b \implies ab=1 \implies k=1,$ a contradiction. Therefore $f(x)=x$ for all $x,$ which is a valid solution. Hence, we have determined that the only solutions to the functional equation are $f(x)=x, f(x)=\frac{1}{x}.$ It is easy to verify that they work.