IMO 2008, Question 4

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3647 posts

orl

#1 h Jul 17, 2008, 12:45 PM • 16 Y

Y by Davi-8191, integrated_JRC, Vietjung, HWenslawski, Adventure10, megarnie, mathmax12, jmiao, aidan0626, Amir Hossein, buddyram, ItsBesi, cubres, and 3 other users

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 numbers $w,x,y,z,$ satisfying $wx = yz.$

Author: Hojoo Lee, South Korea

This post has been edited 4 times. Last edited by MellowMelon, Sep 3, 2015, 12:05 AM
Reason: fix typo

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the.sceth

16 posts

the.sceth

#2 h Jul 17, 2008, 1:28 PM • 7 Y

Y by Adventure10, HoRI_DA_GRe8, Mango247, Amir Hossein, and 3 other users

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.

This post has been edited 1 time. Last edited by the.sceth, Jul 17, 2008, 9:55 PM

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not_trig

2049 posts

not_trig

#3 h Jul 17, 2008, 1:36 PM • 4 Y

Y by Plops, Adventure10, Mango247, and 1 other user

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 $p,q,s$ ?

This post has been edited 1 time. Last edited by not_trig, Jul 17, 2008, 1:38 PM

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TTsphn

1313 posts

TTsphn

#4 h Jul 17, 2008, 1:37 PM • 10 Y

Y by lordnmt, YadisBeles, Imayormaynotknowcalculus, Adventure10, myh2910, Mango247, Wunka, and 3 other users

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) $f(x)\equiv \frac {1}{x}$

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lambruscokid

264 posts

lambruscokid

#5 h Jul 17, 2008, 2:56 PM • 2 Y

Y by Adventure10, Mango247

My solution (at home) is very similar to TTsphn's one...but quite longer and messier

By the way nice solution!

Daniel

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Ibn Rochde

88 posts

Ibn Rochde

#6 h Jul 17, 2008, 3:46 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

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 $f(z)=(1/z)$

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Lepuslapis

78 posts

Lepuslapis

#7 h Jul 17, 2008, 4:07 PM • 2 Y

Y by Adventure10, Mango247

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).

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Ibn Rochde

88 posts

Ibn Rochde

#8 h Jul 17, 2008, 4:54 PM • 2 Y

Y by Adventure10, Mango247

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 $x \in [a.+00[$

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the.sceth

16 posts

the.sceth

#9 h Jul 17, 2008, 5:21 PM • 2 Y

Y by Adventure10, Mango247

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.

This post has been edited 1 time. Last edited by the.sceth, Jul 17, 2008, 9:54 PM

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campos

411 posts

campos

#10 h Jul 17, 2008, 5:38 PM • 2 Y

Y by Adventure10 and 1 other user

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...

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giove

13 posts

giove

#11 h Jul 17, 2008, 9:58 PM • 4 Y

Y by Adventure10, Mango247, and 2 other users

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.

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jasamper88

187 posts

jasamper88

#12 h Jul 17, 2008, 11:01 PM • 1 Y

Y by Adventure10

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 $\displaystyle f(ab)= \frac {ab + a +\frac 1b - b}{a+b}<0$ ).

This post has been edited 1 time. Last edited by jasamper88, Jul 18, 2008, 7:09 AM

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me@home

2349 posts

me@home

#13 h Jul 18, 2008, 4:46 AM • 5 Y

Y by mostafataheri, ArefS, green_leaf, Adventure10, and 1 other user

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

1)
2)
3)

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gauss_chen

2 posts

gauss_chen

#14 h Jul 21, 2008, 8:37 AM • 2 Y

Y by Adventure10, Mango247

This problem seems not very hard

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t0rajir0u

12167 posts

t0rajir0u

#15 h Jul 22, 2008, 10:04 AM • 3 Y

Y by Adventure10, Mango247, Spiritpalm

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.

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ryanbear

1056 posts

ryanbear

#110 h Feb 12, 2024, 5:03 AM • 2 Y

Y by Spiritpalm, Amir Hossein

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

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eibc

600 posts

eibc

#111 h Mar 10, 2024, 11:50 PM • 1 Y

Y by Amir Hossein

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.

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peppapig_

281 posts

peppapig_

#112 h Mar 31, 2024, 4:11 PM • 2 Y

Y by Amir Hossein, Jndd

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.

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KevinYang2.71

421 posts

KevinYang2.71

#113 h Apr 26, 2024, 6:48 AM • 3 Y

Y by Amir Hossein, megarnie, deduck

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. $\square$

This post has been edited 1 time. Last edited by KevinYang2.71, Apr 26, 2024, 6:49 AM

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Mr.Sharkman

498 posts

Mr.Sharkman

#114 h May 27, 2024, 10:40 PM

Y by

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.

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Jndd

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Jndd

#115 h Jul 26, 2024, 3:47 PM

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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.

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ezpotd

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ezpotd

#116 h Aug 30, 2024, 8:39 PM

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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 $wx = yz = y^2 \neq 1$ .

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pie854

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pie854

#117 h Sep 10, 2024, 5:00 AM

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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.

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bjump

1015 posts

bjump

#118 h Oct 28, 2024, 7:31 PM

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Solution

This post has been edited 1 time. Last edited by bjump, Oct 28, 2024, 7:31 PM

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Siddharthmaybe

106 posts

Siddharthmaybe

#119 h Dec 30, 2024, 8:37 PM

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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+
Do Pointwise trap obviously otherwise ur stupid
both of which can be easily verified

This post has been edited 1 time. Last edited by Siddharthmaybe, Apr 3, 2025, 7:37 PM
Reason: ptwise

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Maximilian113

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Maximilian113

#120 h Jan 5, 2025, 11:59 PM

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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.

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blueprimes

351 posts

blueprimes

#122 h Feb 27, 2025, 2:32 AM

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imma crash out if i have to bash another pointwise trap

We claim the solutions are $f(w) \equiv w, \dfrac{1}{w}$ which can easily be checked to work. Now we prove these are the only ones, setting $x = z = 1$ we obtain $f(w)^2 + f(1)^2 = f(w^2) + f(1) \implies f(w)^2 = f(w^2) + c$ for some constant $c$ . Substituting back into the given equation and replacing $w^2 \mapsto w$ and vice versa we obtain
$\dfrac{f(w) + f(x) + 2c}{f(y) + f(z)} = \dfrac{w + x}{y + z}$ for all $wx = yz$ . Let the above assertion be $P(w, x, y, z)$ . Now $P(w, w, w, w)$ gives $c = 0$ then $f(1) - f(1)^2 = 0 \implies f(1) = 1$ and $f(w^2) = f(w)^2$ for all $w$ . Now $P(w^2, 1, w, w)$ yields
$\dfrac{f(w^2) + f(1)}{2 f(w)} = \dfrac{f(w)^2 + 1}{2 f(w)} = \dfrac{w^2 + 1}{2w} \implies f(w) \in \left \{w, \dfrac{1}{w} \right \} \forall w.$ From here we solve the pointwise trap, for the sake of contradiction assume $a, b \ne 1$ exist where $f(a) = a, f(b) = \dfrac{1}{b}$ . Then $P(a^2, b^2, ab, ab)$ yields
$\dfrac{f(a^2) + f(b^2)}{2 f(ab)} = \dfrac{a^2 + b^2}{2ab} \implies a(a^2b^2 + 1) = b(a^2 + b^2) f(ab)$ from $f(a^2) = a^2$ and $f(b^2) = \dfrac{1}{b^2}$ . If $f(ab) = ab$ , the above relation simplifies to
$a(a^2b^2 + 1) = ab^2(a^2 + b^2) \implies a^3b^2 + a = a^3b^2 + ab^4 \implies b^4 = 1 \implies b = 1$ which fails. On the other hand, if $f(ab) = \dfrac{1}{ab}$ we obtain
$a^2b(a^2b^2 + 1) = b(a^2 + b^2) \implies a^4b^2 + a^2 = a^2 + b^2 \implies a^4 = 1 \implies a = 1$ which also fails. Contradiction, so $f(w) \equiv w, \dfrac{1}{w}$ are our only solutions as wanted.

This post has been edited 1 time. Last edited by blueprimes, Feb 27, 2025, 2:36 AM

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Marcus_Zhang

980 posts

Marcus_Zhang

#123 h Mar 15, 2025, 1:30 AM

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FE with quadratics??

This post has been edited 1 time. Last edited by Marcus_Zhang, Mar 15, 2025, 1:31 AM

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Ilikeminecraft

615 posts

Ilikeminecraft

#124 h Apr 25, 2025, 7:59 AM

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I claim the answer is $f(x) = x$ or $f(x) = \frac1x.$

First, let $w = x = y = z = 1.$ We see that $f(1)^2= f(1),$ and thus $f(1) = 1$ because the codomain is positive reals. Next, let $w = x = 1, yz = 1.$ Hence, we have that $f(y^2) + f\left(\frac1{y^2}\right) = y^2 + \frac1{y^2}.$ However, since our function is over the positive reals, we can assume that $f(x) + f\left(\frac1x\right) = x +\frac1x.$ Now, we plug in $wx = 1, y = z = 1.$ Hence, $f(x)^2 + f\left(\frac1x\right)^2 = x^2 + \frac1{x^2}.$ By squaring the old equation and subtracting these two, we see that $f(x)f\left(\frac1x\right)= 1.$ Now, by solving this with the other equation, we see that $f(x) = x, \frac1x.$ To address pointwise, we can just trivially do casework to show that it can only be one.

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lpieleanu

2994 posts

lpieleanu

#125 h Yesterday at 4:11 PM

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Solution

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