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267 posts

#1 h May 8, 2022, 5:01 PM • 1 Y

Y by Sedro

Find all functions $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$ such that
$f(xf(x + y)) = yf(x) + 1$ holds for all $x, y \in \mathbb{R}^{+}$ .

Proposed by Nikola Velov, North Macedonia

This post has been edited 2 times. Last edited by Lukaluce, May 10, 2022, 2:31 PM

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#2 h May 8, 2022, 8:47 PM • 3 Y

Y by tiendung2006, XcFL, Sedro

solution

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1284 posts

#3 h May 9, 2022, 8:22 AM

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Clearly $f$ is surjective on $(1,\infty)$ and injective as if $f(z_1) = f(z_2)$ , then $y=z_1-x$ and $y=z_2-x$ for any $x<z_1,z_2$ yield $f(xf(z_1)) = (z_1-x)f(x) + 1$ , $f(xf(z_2)) = (z_2-x)f(x) + 1$ and hence $z_1=z_2$ . To use these properly, take any $z>1$ , consider $x_0$ with $z=f(x_0)$ and aim to make the right-hand side of the initial equation with an $f$ in order to use injectivity. To be precise, setting $x=x_0$ and $y=\frac{z-1}{z}$ yields $f(x_0f(x_0+\frac{z-1}{z})) = f(x_0)$ , thus $f(x_0 + \frac{z-1}{z}) = 1$ and so $x_0 + \frac{z-1}{z} = k$ for some constant $k$ , thus
$\mbox{If } x \mbox{ is such that }f(x) > 1, \mbox{ then } f(x) = \frac{1}{x+1-k}.$ Now, in the initial equation we have $f(xf(x+y)) = yf(x) + 1 > 1$ , hence this equation becomes (with $f(k) = 1$ )
$\frac{1}{xf(x+y) + 1-k} = yf(x) + 1$ Setting $y=k-x$ yields $1 = (x+1-k)((k-x)f(x) + 1) \Leftrightarrow f(x) = \frac{1}{x+1-k}$ for all $x<k$ . Setting $x=k$ yields $\frac{1}{kf(k+y) + 1 -k} = y+1$ and hence $kf(z) + 1 - k = \frac{1}{z-k+1} \Leftrightarrow f(z) = \frac{(k-1)z-k^2+2k}{zk - k^2+k}$ for all $z>k$ . Now from $\frac{1}{x+1-k} = f(x) > 0$ we have $x > k-1$ for all $0<x<k$ , thus $k\leq 1$ ; and in $\frac{(k-1)z-k^2+2k}{zk - k^2+k} = f(z) > 0$ for large $z$ and fixed $k<1$ the numerator will be negative and the denominator will be positive, contradiction! Therefore $k=1$ and substituting above yields $f(x) = \frac{1}{x}$ for all of $x<1$ , $x=1$ and $x>1$ , i.e. $f(x) = \frac{1}{x}$ for all $x$ (which satisfies the given equation).

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#5 h May 15, 2022, 8:24 AM • 1 Y

Y by math_comb01

Very nice

we prove that the only function which works is $f(x)=\dfrac 1x$ and it is easy to verify that this indeed works.

Let $P(x,y)$ denote the given assertion. Firstly, fixing $x$ gives that $f$ is surjective over $(1, \infty)$ . If $f(a)=f(b)$ , then pick $c < \min\{a,b\}$ , equating $P(c,a-c)$ and $P(c,b-c)$ gives $a=b$ hence $f$ is injective.

Claim:
$f\left(x + \dfrac{1}{f(x)}\right) = \dfrac cx$ for some constant $c$ also from here onwards let this assertion denote $Q(x)$ .
Proof: Equating $P(1,1/f(1))$ and $P(x, 1/f(x))$ we have
$f(xf(x + 1/f(x)))=2=f(f(1+1/f(1)))$ and because of injectivity, $xf(x+1/f(x)) = f(1 + 1/f(1)) = c$ .

Claim: $f(1)=1$ .
Proof: $Q(c) \implies$ there exists $t$ such that $f(t)=1$ . If $t > 1$ then $P(1,t-1) \implies f(t-1) = f(t-1) + 1$ which is a contradiction, hence $t \le 1$ . Note that $Q(x)$ means that $f$ is completely surjective, if $t <1$ then choose $x_0$ such that $f(x_0)=\dfrac{1}{1-t}$ then $P(x_0,t) \implies f(x_0f(x_0 + t)) = tf(x_0)+1 = \dfrac{1}{1-t} = f(x_0)$ and because of injectivity it means $f(x_0 + t) = 1 = f(t)$ which is a contradiction. To conclude, $f(1)=1$ .

Note that $P(x,1-x) \implies f(x) = \dfrac 1x$ for all $x < 1$ . From $f(1)=1$ , $Q(1)\implies c=f(2)$ , $P(1,1) \implies f(c) = f(f(2))=2$ , $P(1/2 , 1/2) \implies f(1/2) = 2 = f(c) \implies c=1/2$ . $P(1,x) \implies f(f(1+x))=1+x \implies f(f(x)) = x$ for all $x\ge 1$ .

Claim: $f(x) = \dfrac 1x$ for all $x \ge 1$ .
Proof: Choose $x\ge 1$ ,
$f(Q(x)) \implies x + \dfrac{1}{f(x)} = f(f(x + 1/f(x))) = f\left(\dfrac{1}{2x}\right)$ and since $x\ge 1 \implies \dfrac{1}{2x} <1$ thus
$x + \dfrac{1}{f(x)} = f\left(\dfrac{1}{2x}\right) = 2x \implies f(x) = \dfrac 1x$ and this solves the problem.

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#6 h Jun 15, 2022, 10:28 PM

Y by

The answer is $f(x) \equiv \frac{1}{x}$ . It is easy to see that it works.

Let $P(x,y)$ be the given assertion.

Claim 1: $f$ is injective.

Proof: Suppose $f(a) = f(b)$ . Fix a small $x$ (we only require $x < a,b$ ). Comparing $P(x,a-x)$ and $P(x,b-x)$ gives
$(x-a)f(x) = (x-b)f(x)$ As $f(x) \ne 0$ , hence $a=b$ , as desired. $\square$

Call two pairs $(x_1,y_1)$ and $(x_2,y_2)$ similar (and denote $(x_1,y_1) \sim (x_2,y_2)$ ) if
$y_1f(x_1) = y_2f(x_2) \qquad \qquad (1)$ Comparing $P(x_1,y_1)$ and $P(x_2,y_2)$ (and invoking $f$ is injective) gives
$x_1f(x_1 + y_1) = x_2 f(x_2 + y_2)\qquad \qquad (2)$ whenever $(x_1,y_1) \sim (x_2,y_2)$ . Basically, if $(1)$ is true $\iff$ $(2)$ is true. Note in some sense $(1)$ is just fixing the ratio of $y_1,y_2$ .

Claim 2: $f$ is strictly decreasing.

Proof: Recall $f$ was injective. Assume on the contrary that $x_1 < x_2 ~~ \text{and} ~~f(x_1) < f(x_2)$ Consider all $y_1,y_2$ such that $(1)$ is true. Then we have
$x_1f(x_1 + y_1) = x_2 f(x_2 + y_2)$ But due to our assumptions, $y_1,y_2$ can be chosen such that
$x_1 + y_2 = x_2 + y_2$ So that forces
$x_1 = x_2$ which is a contradiction. $\square$

Claim 3: $f$ is continuous.

Proof: (We basically vary $y$ in $P(x,y)$ to prove this) Fix $x$ and consider any $a$ . It suffices to show (as we could take $x$ small) that
$f(x+a) = \lim_{z \to a} f(x+z)$ Let $I$ be any of the below two sets (or intervals)
$\{z : 0 < a < x+a \} ~~,~~ \{z : a < z\}$ Pick any small $\epsilon > 0$ . As $f$ is strict monotone, so it suffices to show set
$f(x + I) - f(x+a)$ has an element $\in (-\epsilon,\epsilon)$ . Assume on the contrary that
$|f(x+z) - f(x+a)| \ge \epsilon ~~ \forall ~ z \in I$ So there is a $\alpha > 0$ such that
$|xf(x+a) - xf(x+z)| \ge \alpha ~~ \forall ~ z \in I$ As $f$ is strict monotone, so $\exists ~ \beta >0$ such that
$\left\vert f(xf(x+a)) - f(xf(x+z)) \right\vert \ge \beta ~~ \forall ~ z \in I$ But then invoking $P(x,a)$ and $P(x,z)$ gives
$\left\vert af(x) - zf(x) \right\vert \ge \beta ~~ \forall ~ z \in I$ But tending $z \to a$ , the LHS tends to $0$ , so we obtain our desired contradiction. $\square$

Look at $(1),(2)$ now. Fix $x_1,x_2$ . Pick $y_1,y_2$ satisfying $(1)$ and tend $y_1,y_2 \to 0$ . Using continuity of $f$ we obtain
$x_1f(x_1) = x_2f(x_2) ~~ \forall ~ x_1,x_2$ It follows there exists a $c$ such that
$f(x) \equiv \frac{c}{x}$ Now there are several ways to conclude $c=1$ :

Just plug in $f$ in $P(x,y)$ .
Note $f$ is always a involution. $P(1,y)$ gives
$yf(1) + 1 = f(f(1+y)) = 1+y$ Which forces $f(1) = 1$ , hence $c=1$ .

This completes the proof. $\blacksquare$

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ZETA_in_olympiad

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ZETA_in_olympiad

#7 h Jun 22, 2022, 9:22 AM

Y by

We claim that $f\equiv \text{Id}^{-1}$ , which clearly works. Denote the assertion by $P(x,y).$

Claim: $f$ is bijective.
Proof. For injectivity, take $f(u)=f(v)$ and a small $w$ then by combining $P(w,u-w)$ and $P(w,v-w)$ we get $u=v.$ Then take $P(x,\tfrac{1}{f(x)})$ so that we get $xf(x+\tfrac{1}{x})$ constant. This is enough for surjectivity. $\blacksquare$

Pick an $a$ for $f(a)=1.$ If $a>1$ then $P(a-1,1)$ gives contradiction. If $a<1$ then take $b$ for $f(\tfrac{1}{b})=1-a.$ By $P(\tfrac{1}{b},a)$ we get contradiction. Note that $f(1)=1.$

For all $x>1$ by $P(1,x-1)$ and for all $x<1$ by $P(x,1-x)$ yields $f(x)=\tfrac{1}{x},$ as desired.

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Iora

#8 h Sep 3, 2022, 2:05 AM

Y by

I hope it works
Let $P(x,y)$ be the assertion of the given functional equation.

It is not so hard to prove $f$ is bijective. Without a problem, let $y \rightarrow 0$ . We have:
$f(xf(x))=1$ Obviously, $f$ is not constant. Hence we can conclude that $xf(x)$ is constant. Therefore, $f(x)=\frac{k}{x}$ for some positive real number $k$ . Putting to the our equation, we get:

$\frac{x+y}{x}= \frac{ky+x}{x}$
Hence we conclude $k=1$ which gives $f(x)=\frac{1}{x} \ \blacksquare$

Note:Solution is wrong since continuity of the function is not proven, hence we can't take $y \rightarrow 0$

This post has been edited 1 time. Last edited by Iora, Sep 12, 2022, 12:10 PM

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#9 h Dec 11, 2022, 10:23 AM

Y by

Because of the original equation there exist an $x_0$ such that $f(x_0)>1$
As solutions above, we can show $f$ is injective.
$P(x_0,\frac{f(x_0)-1}{f(x_0)})$ gives $f(x_0+ \frac{f(x_0)-1}{f(x_0)}) = 1$ . So there is an $x_1$ such that $f(x_1)=1$

$P(x,x_1-x),(x<x_1)$ gives $f(x)= \frac{1}{x+1-x_1}$
Plugging infinite $x,y$ pairs such that $x+y<x_1$ and $yx_1>x$ gives $x_1 = 1$ and $f(x)=x$ for $x<1$
$P(1,y)$ gives $f(f(x))=x$ for $x>1$ , but $f(\frac{1}{x})=x= f(f(x))$ and from injectivity $f(x)=\frac{1}{x}$ for $x>1$
Hence, $f(x)= \frac{1}{x}$ for all $x>0$

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ahizounemustapha

68 posts

ahizounemustapha

#10 h May 20, 2023, 10:45 PM

Y by

we note h(χ) the function such that if x>0 so h(x)=f(x)
and h(x)=0 if x=0
proof : h(xh(x+y))=yh(x)+1
p(0,y): h(0)=yh(0)+1
so if h(0)≠0 y=1/h(0)+1 which is a point fix a contra diction because y is free over R+
so h(0)=0
p(x,-x);remark h is defind over R
so h(0)=-xh(x)+1
then h(x)=1/x
so over R+ :f(x)=1/x

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#11 h Nov 13, 2023, 5:32 PM • 1 Y

Y by OronSH

Solved with OronSH and pi271828.

The only solution is $\boxed{f(x) = \frac{1}{x} }$ . To see this works, note the left hand side becomes $\frac{x+y}{x}$ and the right hand side becomes $\frac{y}{x} + 1 = \frac{x+y}{x}$ . Now we prove it's the only solution.

Claim: $f$ is injective.
Proof: Suppose $f(a) = f(b)$ for some positive reals $a,b$ . Fix an $0<x < \min(a,b)$ .

$P(x, a-x): f(x f(a)) = (a-x) f(x) + 1$ .

$P(x,b-x): f(xf(a)) = (b-x) f(x) + 1$ .

So $(a-x)f(x) + 1 = (b-x) f(x) + 1\implies a = b$ . $\square$

$P\left( x, \frac{1}{f(x)} \right) : f\left( x f\left(x + \frac{1}{ f(x)} \right)\right) = 2\implies f\left( x + \frac{1}{f(x)} \right) = \frac{k}{x} \ \ \ \ \ \ \ \ (1)$ where $k$ is the unique positive number with $f(k) = 2$ .

The RHS of $(1)$ takes all positive real numbers, so $f$ is surjective. Now plugging $x = k$ in $(1)$ gives that $f\left( k + \frac{1}{2} \right) = 1$ .

For $x < k + \frac{1}{2}$ , $P\left( x, y = k + \frac{1}{2} - x \right):f(x) = yf(x) + 1$ , so $f(x) = \frac{1}{1-y} = \frac{1} {x + \frac{1}{2} - k }$ This implies that $f\left( \frac{k}{2} \right) = \frac{2}{1 - k}$ .

For any $x < k$ , $P(x, k - x): f(2x) = (k-x) f(x) + 1$ . Setting $x = \frac{k}{2}$ gives that $2 = \frac{k}{2} f\left( \frac{k}{2} \right) + 1,$ so $f\left( \frac{k}{2} \right) = \frac{2}{k}$ , which means $\frac{2}{k} = \frac{2}{1-k} \implies k = \frac{1}{2}$
Then we have $f(x) = \frac{1}{x}$ for all $x<1$ .

Now since $f\left( x + \frac{1}{f(x)} \right) = \frac{1/2}{x}$ , if $f(x) = \frac{1}{x}$ , then $f(2x) = \frac{1}{2x}$ , which means if all $x\in (a,b)$ satisfy $f(x) = \frac{1}{x}$ , then all $x$ in $(2a, 2b)$ satisfy $f(x) = \frac{1}{x}$ .

Since all $x\in (0,1)$ satisfy $f(x) = \frac{1}{x}$ , we can induct to get that all $x\in (0, 2^n)$ for any nonnegative integer $n$ satisfy $f(x) = \frac{1}{x}$ . Since $2^n$ is unbounded, all positive reals $x$ satisfy $f(x) = \frac{1}{x}$ .

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ATGY

#12 h Jan 26, 2024, 12:09 PM

Y by

Nice! Let $P(x, y)$ be the given assertion. Firstly, we claim that the function is injective. Let $f(a) = f(b)$ . $P(x, a - x)$ , $P(x, b - x)$ gives:
$af(x) + 1 = bf(x) + 1 \implies a = b$ After substituting $P(x, \frac{1}{f(x)})$ , we see that the function is surjective. Now, let $f(\alpha) = 1$ . $P(1, y)$ yields:
$f(f(y + 1)) = yf(1) + 1$ Plugging in $P(f(y) + 1, \frac{yf(1)}{yf(1) + 1})$ , we get:
$f(f(y+1)f(f(y+1) + \frac{yf(1)}{yf(1) + 1})) = \frac{yf(1)}{yf(1) + 1)}\times f(f(y+1)) + 1 = yf(1) + 1$ By injectivity of $f(x)$ , we have:
$f(f(y+1)f(f(y+1) + \frac{yf(1)}{yf(1) + 1})) = f(f(y + 1)) \implies f(y+1)f(f(y+1) + \frac{yf(1)}{yf(1) + 1} = f(y + 1)$ $\implies f(y + 1) + \frac{yf(1)}{yf(1) + 1} = 1$ Since $f(\alpha) = 1$ , we have:
$\alpha - \frac{yf(1)}{yf(1) + 1} = f(y + 1) \; \; \; \; (1)$ For sufficiently large $y$ , $\frac{yf(1)}{yf(1) + 1}$ tends to $1$ (but not exactly 1), and since $f(y + 1)$ is positive we have $\alpha \geq 1$ .

We now claim that $f(1) = 1$ . Say, for the sake of contradiction, $\alpha > 1$ . $P(\alpha - 1, 1)$ gives:
$f(\alpha - 1) = f(\alpha - 1) + 1$ This is clearly not possible which means that $\alpha = 1$ . Substituting this value in $(1)$ , we have:
$f(y + 1) = \frac{1}{y + 1}$ For $y \geq 1$ , we have $f(y) = \frac{1}{y}$ . We prove this for $y < 1$ as well. $P(y, 1 - y)$ gives:
$f(y) = (1 - y)f(y) + 1 \implies yf(y) = 1 \implies f(y) = \frac{1}{y}$ Hence, $f(x) = \frac{1}{x}$

This post has been edited 2 times. Last edited by ATGY, Feb 6, 2024, 12:45 PM

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#13 h Feb 5, 2024, 6:09 PM

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ZETA_in_olympiad wrote:

Pick an $a$ for $f(a)=1.$ If $a>1$ then $P(a-1,1)$ gives contradiction. If $a<1$ then take $b$ for $f(\tfrac{1}{b})=1-a.$ By $P(\tfrac{1}{b},a)$ we get contradiction. Note that $f(1)=1.$

Can you please clarify how you get contradiction for $a<1$ ?

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Marius_Avion_De_Vanatoare

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Marius_Avion_De_Vanatoare

#14 h Apr 8, 2024, 7:01 AM

Y by

As usual let $P(x;y)$ denote the given assertion.
Assume the function is not injective and let $f(a)=f(b)$ than setting $x+y_1=a$ and $x+y_2=b$ $P(x;y_1)-P(x;y_2) \Rightarrow (a-b)f(x)=0$ , a contradiction.
Now $P(x;\frac{y}{f(x)}) \Rightarrow f(xf(x+\frac{y}{f(x)}))=y+1$ , in particular the expression inside only depends in $y$ , not in $x$ , so let $xf(x+\frac{y}{f(x)})=g(y)$ , and setting $x$ to be $g(y)$ we get that there is a value $\alpha$ such that $f(\alpha)=1$ . Take $x+y=\alpha$ and we get that for any $x<\alpha, f(x)=\frac{1}{x+1-\alpha}$ and now we know the form of $f$ for sufficiently small inputs. Fixing $x+y<\alpha$ and taking $x$ very very small we get $\alpha=1$ and now again fixing $x+y$ but now for arbitrary values and taking $x\rightarrow 0$ we get that $f(a)=\frac{1}{a}$ for any positive real $a$ which is indeed a solution.

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mcmp

#15 h Oct 14, 2024, 10:56 AM • 1 Y

Y by ohiorizzler1434

Wooahhhhhh beautiful and crazy problem

First analysis-style functional equation that I’ve solved myself

yay

This is going to be a LOT different to the solutions above :whistling:

:whistling:

I don’t know how I managed to cook this up.

So firstly we claim $f(x)=\frac{1}{x}$ is a working function, which is easy checking. Now we check that the related functions $f(x)=\frac{c}{x}$ for $c\neq1$ don’t work. Indeed:
$\begin{align*} f(xf(x+y))&=f\left(x\frac{c}{x+y}\right)\\ &=\frac{x+y}{x}\\ &=\frac{y}{x}+1\\ yf(x)+1&=\frac{cy}{x}+1 \end{align*}$ so we’re forced to have $c=1$ . So it suffices now to show that $f(x)=\frac{c}{x}$ for some constant $c$ as per the above.

We now split the proof into 5 (!) parts, the first three of which are easy and elementary, whilst the last two really start to come scarily close to real analysis :noo:

:noo:

Part 1: $f$ is injective and surjective over all real numbers $>1$
Assume $f(a)=f(b)$ . Choose $c<a,b$ , and $A=a-c$ , $B=b-c$ .
$\begin{align*} f(a)&=f(b)\\ f(c+A)&=f(c+B)\\ f(cf(c+A))&=f(cf(c+B))\\ Af(c)+1&=Bf(c)+1\\ A&=B\\ a&=b \end{align*}$ Also note by varying $y>0$ that $f(xf(x+y))$ attains any positive real number $>1$ .

Part 2: $af(a+yf(b))=bf(b+yf(a))$ for any two $a,b\in\mathbb{R}_{>0}$
Notice both are just equal to $f^{-1}(yf(a)f(b)+1)$ (here the inverse is well defined as clearly from the functional equation some $c\in\mathbb{R}_{>0}$ satisfies $f(c)=yf(a)f(b)+1$ , and since $f$ is injective this $c$ is unique).

Part 3: $f$ strictly decreasing
Take $a<b$ . Note $f(a)\neq f(b)$ so assume FTSoC $f(a)<f(b)$ . Now consider a $y\in\mathbb{R}_{>0}$ such that $a+yf(b)=b+yf(a)$ (this actually can be explicitly computed! $y=\frac{b-a}{f(b)-f(a)}>0$ by assumption). Hence $af(b+yf(a))=af(a+yf(b))=bf(b+yf(a))$ , so $a=b$ , contradiction.

Part 4: $\lim_{x\to\infty}f(x)=0$
Notice that $f(x)$ is a strictly decreasing function. Since it is bounded below by $0$ , we can safely claim that $\lim_{x\to\infty}f(x)$ is well behaved! In fact it exists! :harhar:

:harhar:

Call it $A$ . First we do some scouting. Notice that since on the RHS $y$ is free to move around we have some degree of freedom. This calls for some application of inequalities :dry:

:dry:

:censored:

$\begin{align*} f(xf(x+y))&=yf(x)+1\ge Ay+1\\ xf(x+y)&\ge xA>0\\ f(xf(x+y))&\le f(xA)\\ f(xA)&\ge Ay+1 \end{align*}$ So if we send $y\to\infty$ , the fact that $A>0$ implies $Ay+1\to\infty$ . This implies $f(xA)\ge f(xA)+1$ in particular, contradiction. Hence $A=0$ and $\lim_{x\to\infty}f(x)=0$ .
(Only just realised this part was actually redundant oops :huuh:

:huuh:

keeping this still because ~~4 is an unlucky number~~)

Part 5: Finishing up with chains and more inequalities :bomb:

:bomb:

First we pick any $a<b$ (!). In the style of APMO 2023/4 let $(y_n)_{n=-\infty}^{\infty}$ be such that $y_0=1$ and $b+f(a)y_{n-1}=a+f(b)y_n$ . Let $x_n=a+f(b)y_n$ . Note now $af(x_n)=bf(b+f(a)y_n)=bf(a+f(b)y_n)=bf(x_{n+1})$ , so $f(x_{n-1})=\frac{b}{a}f(x_n)>f(x_n)$ . Hence $x_{i}$ is strictly increasing. Notice that $f(x_{i})=\frac{b^i}{a^i}f(x_0)=\left(\frac{b}{a}\right)^if(x_0)$ so $x_{-i}\to0$ as $i\to\infty$ . The reason for this is as follows: first notice that from $f$ surjective on real numbers $>1$ that $\exists c\in\mathbb{R}_{>0}$ such that $f(c)=r$ for any $r>1$ . Now since $(x_{-n})_{n=0}^{\infty}$ is strictly decreasing and bounded from below by $0$ , $c=\lim_{n\to\infty}x_{-n}$ exists again by the monotone convergence theorem. Hence noticing that if $x_{-i}\to c$ for some $c>0$ , there are still $0<d<c$ , so considering $f(d)$ , $f(x_{-i})<f(d)$ , hence $\left(\frac{b}{a}\right)^nf(x_0)<f(d)$ for all $n>0$ , a clear contradiction since $\frac{b}{a}>1$ .

Hence $\lim_{n\to\infty}x_{-n}=0$ . But then this implies quite quickly that $\lim_{n\to\infty}y_{-n}=-\frac{a}{f(b)}$ . However we can compute $y_{-n}$ quite explicitly. Noticing that $y_{n-1}=\frac{a-b+f(b)y_n}{f(a)}$ , we can quite quickly deduce from induction that for $n\ge1$ :
$\begin{align*} y_{-n}&=\frac{a-b}{f(a)}\left(\sum_{i=0}^{n-1}\left(\frac{f(b)}{f(a)}\right)^i\right)+\frac{f(b)^{n}}{f(a)^{n}} \end{align*}$ However notice in the limit as $n\to\infty$ , $\frac{f(b)^{n}}{f(a)^{n}}\to0$ , whilst since $f(b)<f(a)$ :
$\begin{align*} \frac{a-b}{f(a)}\left(\sum_{i=0}^{n-1}\left(\frac{f(b)}{f(a)}\right)^i\right)&\to\frac{a-b}{f(a)}\left(\sum_{i=0}^{\infty}\left(\frac{f(b)}{f(a)}\right)^i\right)\\ &=\frac{a-b}{f(a)}\frac{1}{1-\frac{f(b)}{f(a)}}\\ &=\frac{a-b}{f(a)-f(b)} \end{align*}$ So $\lim_{n\to\infty}y_{-n}=\frac{a-b}{f(a)-f(b)}=-\frac{a}{f(b)}$ . Now by a very useful result called addendo, $\frac{a-b-a}{f(a)-f(b)-(-f(b))}=-\frac{a}{f(b)}$ , or $\frac{b}{f(a)}=\frac{a}{f(b)}$ . Hence $af(a)=bf(b)$ . We never really fixed down $a$ and $b$ though, so in fact $af(a)$ is constant which is to say $f(a)=\frac{c}{a}$ for some constant $c$ for all real numbers $a>0$ , ending the proof.

:bomb:

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#16 h Jan 16, 2025, 12:31 PM

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$f(xf(x + y)) = yf(x) + 1$ Only function satisfying the equation is $f(x)=\frac{1}{x}$ .
Claim: $f$ is injective.
Proof: Suppose that $f(a)=c=f(b)$ . Pick $y=a-x$ and then $y=b-x$ (and choose $x$ sufficiently small) in order to see that $(b-x)f(x)+1=f(cx)=(a-x)f(x)+1$ which implies $a=b$ so $f$ is injective. $\square$
Claim: $f$ is surjective.
Proof: Plugging $x,y/f(x)$ gives $f$ is surjective for $>1$ .
$f(xf(x+\frac{y}{f(x)}))=y+1=f(zf(z+\frac{y}{f(z)}))\implies xf(x+\frac{y}{f(x)})=zf(z+\frac{y}{f(z)})$ $z=1$ yields $xf(x+\frac{y}{f(x)})=f(1+\frac{y}{f(1)})$ . Write $xf(1+\frac{y}{f(1)})$ instead of $x$ to see that $f$ is surjective. $\square$
Claim: $f(1)=1$ .
Proof: Let $f(a)=1$ and suppose that $a\neq 1$ . For $x<a$ , we have $f(x)=f(xf(a))=(a-x)f(x)+1$ or $f(x)=\frac{1}{x+1-a}$ . Note that this requires $1\geq a$ hence $1>a$ by our assumption. Pick $x<a(1-a)<a$ and $x+y<a$ to get $f(\frac{x}{x+y+1-a})=\frac{y}{x+1-a}+1=\frac{x+y+1-a}{x+1-a}$ . Since $\frac{x}{x+y+1-a}<a$ holds, we see that $\frac{1}{\frac{x}{x+y+1-a}+1-a}=\frac{x+y+1-a}{x+1-a}$ or $x+(1-a)(x+y+1-a)=x+1-a$ or $(1-a)(x+y-a)=0$ which is impossible. $\square$
Since $f(1)=1$ , we have $f(x)=\frac{1}{x}$ for $x<1$ . This implies $f(x)>1\iff x<1$ by bijectivity. We have $f(xf(x+y))>1$ so $xf(x+y)<1$ . Thus for $x<1$ ,
$\frac{1}{xf(x+y)}=\frac{y}{x}+1=\frac{x+y}{x}\implies f(x+y)=\frac{1}{x+y}$ As desired. $\blacksquare$

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#17 h Feb 12, 2025, 5:23 AM

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Injectivity is clear, and we can also show that function is surjective when $x>1$ .

For the $x$ that $f(x)>1$ , let's take $y=1-\frac{1}{f(x)}$ . $\implies$ $f(x)=\frac{1}{x+1-c}$ where $c$ is the number that $f(c)=1$ . Sine $f(x)>1$ $\implies$ $c>x$ .

Now let's take $x$ such that $c>x$ . Then we have $\frac{1}{x+1-c} > 1$ , by the surjectivity there exist $a$ such that $f(a)=\frac{1}{x+1-c}$ . But we proved that if $f(a)>1$ then $f(a)=\frac{1}{a+1-c}$ . So $a=x$ which implies that for all $x$ less than $c$ we have $f(x)=\frac{1}{x+1-c}$ . (We can say $c \le 1$ now)

Let's take $k$ such that $k> \frac{1}{c}-1$ $\implies$ $\frac{1}{k+1}>\frac{1-c}{k}$ . Now let's take $x$ such that $\frac{1}{k+1}>x>\frac{1-c}{k}$ , and $y=kx+c-1$ . Then we have $y>0$ and $x+y<c$ , which implies that $const=f(\frac{1}{k+1})=\frac{(k+1)x}{x+1-c}$ , which implies if we move $x$ in the interval we will get that $c=1$ .

We proved that for all $x \le 1$ we have that $f(x)=\frac{1}{x}$ . Now let's take $x<1$ $\implies$ $f(xf(x+y))=\frac{x+y}{x}$ , We already know that $f(\frac{x}{x+y})=\frac{x+y}{x}$ so from the injection: $f(x+y)=\frac{1}{x+y}$ , by moving the $y$ we can easily deduce that $f(x)=\frac{1}{x}$ for all $x>1$ , and we are done.

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Sedro

#18 h Apr 16, 2025, 2:20 AM • 1 Y

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Very nice problem.

We claim that the only solution is $f\equiv x^{-1}$ , which clearly works. We now prove that there are no other solutions. Let $P(x,y)$ denote the assertion.

Claim 1: $f$ is injective.

Proof: Suppose that $a$ and $b$ are positive reals such that $f(a)=f(b)=N$ . Let $c < a,b$ be a positive real; comparing $P(c,a-c)$ and $P(c,b-c)$ gives $f(cN) = af(c)+1 = bf(c)+1$ . Hence, $a=b$ . $\blacksquare$

Claim 2: $(1,\infty)$ is a subset of the range of $f$ .

Proof: Consider the expression $yf(x)+1$ ; note that it is always in the range of $f$ by $P(x,y)$ . Fix $x$ ; as $y$ varies over the positive reals, $yf(x)+1$ takes on every positive real value greater than $1$ , as desired. $\blacksquare$

Claim 3: $1$ is in the range of $f$ .

Proof: Let $r>1$ be a positive real number; we know from claims 1 and 2 that there exists a unique positive real $s$ such that $f(s)=r$ . Then, by $P(s, \tfrac{r-1}{r}$ ), we have $f(sf(s+\tfrac{r-1}{r})) = r$ . Since $f(s)=r$ , by the injectivity of $f$ , we have $f(s+\tfrac{r-1}{r}) = 1$ . $\blacksquare$

Claim 4: Let $k$ be the unique positive real number such that $f(k)=1$ . Then $f(x) = (x+1-k)^{-1}$ for all $x\in (0,k)$ .

Proof: By $P(x,k-x)$ , we have $f(x) = (k-x)f(x) + 1$ , and the desired follows. $\blacksquare$

Claim 5: $f(1)=1$ .

Proof: Consider $(x_0,y_0) = (p,k-p-q)$ , where $p$ and $q$ are positive reals such that $p+q< k$ . We have $x_0+y_0 = k-q < k$ , so from claim 4, $f(x_0+y_0) = (x_0+y_0+1-k)^{-1}$ . Then, $f(x_0f(x_0+y_0)) = f(\tfrac{x_0}{x_0+y_0+1-k})$ . We can choose $p$ and $q$ arbitrarily close to $0$ so that $\tfrac{x_0}{x_0+y_0+1-k} = \tfrac{p}{1-q} < k$ . From claim 4, $f(\tfrac{x_0}{x_0+y_0+1-k}) = (\tfrac{x_0}{x_0+y_0+1-k} + 1-k)^{-1}$ . Since $x_0 = p<k$ , $y_0f(x_0)+1 = \tfrac{x_0+y_0+1-k}{x_0+1-k}$ again from claim 4. Thus, $P(x_0,y_0)$ gives us that $(\tfrac{x_0}{x_0+y_0+1-k} + 1-k)^{-1} = \tfrac{x_0+y_0+1-k}{x_0+1-k}.$ This equation can be re-written as $(1-k)(x_0+y_0-k)=0.$ The value of $x_0+y_0$ is not constant over all possible $(x_0,y_0) = (p,k-p-q)$ ; simply take $p$ and $q$ closer to $0$ , for example. Thus, $k=1$ , as desired. $\blacksquare$

Claim 6: $f(x)=x^{-1}$ for all $x\in(0,1]$ .

Proof: This follows directly from claims 4 and 5. $\blacksquare$

Claim 7: $f$ is an involution.

Proof: Clearly, $f$ is an involution on $(0,1]$ , so it suffices to prove that $f(f(x)) = x$ when $x>1$ . By $P(1,x-1)$ , where $x>1$ , we have $f(f(x)) = (x-1) + 1 = x$ , as desired. $\blacksquare$

Claim 8: $f\equiv x^{-1}$ for all $x>1$ .

Proof: We have $f(f(xf(x+y))) = xf(x+y) = f(yf(x)+1)$ for all positive reals $x$ and $y$ . Let $x$ be any positive real, and let $t$ be a positive real number such that $(x+1)t < 1$ ; note $t<1$ . Then, $P(t,tx)$ gives $f(txf(t)+1) = tf(t(x+1))$ . By claim 6, this equation is equivalent to $f(x+1) = (x+1)^{-1}$ , which proves the claim. $\blacksquare$

Finally, combining claims 5, 6, and 8 yields $f(x)=x^{-1}$ for all positive reals $x$ , and we are done. $\blacksquare$

This post has been edited 2 times. Last edited by Sedro, Apr 16, 2025, 2:25 AM

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#21 h Apr 18, 2025, 3:41 PM

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Lukaluce wrote:

Find all functions $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$ such that
$f(xf(x + y)) = yf(x) + 1$ holds for all $x, y \in \mathbb{R}^{+}$ .

Proposed by Nikola Velov, North Macedonia

Let $P(x,y)$ be the assertion $f(xf(x+y))=yf(x)+1$ .

Claim: $f$ is bijective.
Injectivity then surjectivity. Suppose $f(a)=f(b)$ for some $a,b$ , then choose some $c$ with $c<a,b$ , for instance $c=\frac12\min(a,b)$ . We have:
$P(c,a-c)\Rightarrow f(cf(a))=(a-c)f(c)+1$
$P(c,b-c)\Rightarrow f(cf(a))=(b-c)f(c)+1$
and comparing, we get $a=b$ . Using injectivity, we can de-nest the left hand side:
$P\left(x,\frac{f(y)-1}{f(x)}\right)\Rightarrow xf\left(x+\frac{f(y)-1}{f(x)}\right)=y$ if $f(y)>1$
$P\left(x,\frac{f(xy)-1}{f(x)}\right)\Rightarrow f\left(x+\frac{f(xy)-1}{f(x)}\right)=y$ if $f(xy)>1$
and applying this by setting $x$ such that $f(xy)>1$ , we can do:
$P\left(f(x+y),\frac{xf(y)}{f(f(x+y)))}\right)\Rightarrow f\left(f(x+y)+\frac{xf(y)}{f(f(x+y)))}\right)=y$
which proves that $f$ is surjective.

Claim: $f(1)=1$
Let $c=f^{-1}(1)$ .
$P(x,c-x)\Rightarrow f(x)=\frac1{x+1-c}$ for all $x<c$ , and this implies $x+1-c>0$ for $x>0$ which means that $c\le1$ .
Now if $x<y<c$ then $P(x,y-x)$ becomes $f\left(\frac x{y+1-c}\right)=\frac{y+1-c}{x+1-c}$ , so for instance taking $x=\frac c3$ and $y=\frac c2$ we get:
$f\left(\frac{2c}{6-3c}\right)=\frac{6-3c}{6-4c}.$ But since $\frac{2c}{6-3c}<c$ (rearranges to $c<\frac43$ which is true), we may plug this into $f(x)=\frac1{x+1-c}$ to get:
$f\left(\frac{2c}{6-3c}\right)=\frac{6-3c}{6-7c+3c^2},$ and solving, we get that $c=1$ .

Finish:
From the above, $f(x)=\frac1x$ for all $x\le1$ . If there is some $x>1$ for which $f(x)\ge1$ then $\frac1{f(x)}\le1$ so $f\left(\frac1{f(x)}\right)=f(x)$ , so by injectivity $f(x)=\frac1x<1$ , contradiction. That is, the outputs of $f(x)=\frac1x$ for $x\le1$ cover $[1,\infty)$ , so the outputs of $f(x)$ for $x>1$ are constrained to $(0,1)$ . So if $x>1$ we must have $f(x)<1$ .
$P(1,x)\Rightarrow f(f(x))=x$ for $x>1$
Since $f(x)<1$ we have $f(f(x))=\frac1{f(x)}$ for $x>1$ , so we get $f(x)=\frac1x$ for $x>1$ as well. So whether $x<1$ , $x=1$ , or $x>1$ , we must have $\boxed{f(x)=\frac1x}$ for all $x>0$ , which satisfies the equation.

This post has been edited 3 times. Last edited by jasperE3, Apr 19, 2025, 5:56 PM

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