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My hardest algebra ever created (only one solve in the contest)

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Source: Ukraine IMO TST P9

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#1 h Apr 19, 2025, 9:37 PM • 1 Y

Y by bin_sherlo

Find all functions $f: (0, +\infty) \to (0, +\infty)$ for which, for all $x, y > 0$ , the following identity holds:
$f(x) f(yf(x)) + y f(xy) = \frac{f\left(\frac{x}{y}\right)}{y} + \frac{f\left(\frac{y}{x}\right)}{x}$
Proposed by Mykhailo Shtandenko

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#3 h Apr 19, 2025, 10:47 PM

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Answer is $f(x)=c/x$ which fits. Let $f(x)=g(x)/x$ and $P(x,y)$ be the assertion. We have the equation $\frac{g(\frac{yg(x)}{x})}{y}+\frac{g(xy)}{x}=\frac{g(\frac{x}{y})}{x}+\frac{g(\frac{y}{x})}{y}$ .
By replacing $x,xy$ we get
$g(x^2y)+\frac{g(yg(x))}{y}=g(\frac{1}{y})+\frac{g(y)}{y}\iff g(x^2y)-g(\frac{1}{y})=\frac{g(y)-g(yg(x))}{y}$ Claim: $g(x)=g(xg(y)^2)$ .
Proof: First, $y=1/x$ gives $g(\frac{g(x)}{x})=g(\frac{1}{x})$ . By comparing $P(g(x)/x,y)$ and $P(1/x,y)$ we get
$g(\frac{g(x)^2}{x^2}.y)-g(\frac{1}{y})=-\frac{g(yg(\frac{g(x)}{x}))-g(y)}{y}=-\frac{g(yg(\frac{1}{x}))-g(y)}{y}=g(\frac{1}{x^2}.y)-g(\frac{1}{y})$ Replace $x,g(z)^{2m}y$ to get $g(x^2y)-g(1/y)=\frac{g(y)-g(yg(x))}{g(z)^{2m}y}$ and while we change $m$ , $LHS$ doesn't get changed thus, $g(x^2y)=g(1/y)$ or $g(x)=g(y)$ . We see that constant $g$ holds as desired. $\blacksquare$

@below, the solution except the last part was true, thanks for pointing out

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#4 h Apr 19, 2025, 10:57 PM

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Redacted

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#5 h Apr 19, 2025, 11:15 PM

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Make the replacement $f(x)=\frac{g(x)}{x}$ in order to get that $\frac{g \left(\frac{yg(x)}{x} \right)}{y}+\frac{g(xy)}{x}=\frac{g \left(\frac{x}{y} \right)}{x}+\frac{g \left(\frac{y}{x} \right)}{y}$ and denote this with $P(x,y)$ .
Now by $P(x,xy)$ we get that $\frac{g(yg(x))}{y}+g(x^2y)=g \left(\frac{1}{y} \right)+\frac{g(y)}{y}$ and we call this $Q(x,y)$ for simplicity.
Now notice $P(x,1)$ gives that $g \left(\frac{g(x)}{x} \right)=g \left(\frac{1}{x} \right)$ so suppose that there existed some $g(\ell) \ne 1$ then it means $a=\frac{g(\ell)}{\ell} \ne \frac{1}{\ell}=b$ and $g(a)=g(b)$ so from $Q(a,x)-Q(b,x)$ we get $g(a^2x)=g(b^2x)$ and therefore $g(x)=g(cx)$ for all positive reals $x$ (where we pick $c>1$ but you can also have $c<1$ depending on which $a,b$ is bigger but since we have both it doesn't really matter lol) and now from $Q(x,c^ky)$ we have that $\frac{g(yg(x))}{c^ky}+g(x^2y)=g \left(\frac{1}{y} \right)+\frac{g(y)}{c^ky}$ and in this new equation we can take $k \to \infty$ to get that for all positive reals $x,y$ we have $g(x^2y)=g \left(\frac{1}{y} \right)$ and thus by fixing $y$ and moving $x$ we get that $g$ is constant which just works so either $g(x)=1$ for all positive reals $x$ or $g(x)=d$ for some $d \ne 1$ for all positive reals $x$ and therefore either way $f(x)=\frac{c}{x}$ for all positive reals $x$ thus we are done :cool:

:cool:

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#6 h Apr 19, 2025, 11:23 PM

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mshtand1 wrote:

Find all functions $f: (0, +\infty) \to (0, +\infty)$ for which, for all $x, y > 0$ , the following identity holds:
$f(x) f(yf(x)) + y f(xy) = \frac{f\left(\frac{x}{y}\right)}{y} + \frac{f\left(\frac{y}{x}\right)}{x}$
Proposed by Mykhailo Shtandenko

Note that by plugging $y\mapsto xy$ we get (after clearing denominators):
$xyf(x)f(xyf(x))+x^2y^2f\left(x^2y\right)=f\left(\frac1y\right)+yf(y).$ Let $g(x)=xf(x)$ , and $P(x,y)$ the new assertion $g(yg(x))+yg\left(x^2y\right)=yg\left(\frac1y\right)+g(y)$ .

If $g$ is injective:
$P\left(x,\frac1x\right)\Rightarrow g\left(\frac{g(x)}x\right)=g\left(\frac1x\right)$ , so $g(x)=1$ for all $x$ , contradiction.
So there exist $a\ne b$ with $g(a)=g(b)$ .

$P(a,x)\Rightarrow g(xg(a))+xg\left(a^2x\right)=xg\left(\frac1x\right)+g(x)$
$P(b,x)\Rightarrow g(xg(a))+xg\left(b^2x\right)=xg\left(\frac1x\right)+g(x)$
Comparing, we get $g(px)=g(x)$ where $p=\frac{a^2}{b^2}\ne1$ .
$P(x,py)\Rightarrow g(yg(x))+pyg\left(x^2y\right)=pyg\left(\frac1y\right)+g(y)$
$P(x,y)\Rightarrow g(yg(x))+yg\left(x^2y\right)=yg\left(\frac1y\right)+g(y)$
Comparing, we get $g\left(x^2y\right)=g\left(\frac1y\right)$ , and so (setting $y=1$ ) $g$ must be constant. Therefore $\boxed{f(x)=\frac cx}$ for all $x>0$ , which satisfies the equation for any constant $c>0$ .

fakesolve

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#7 h Apr 19, 2025, 11:36 PM • 1 Y

Y by truongphatt2668

Same as above, we use the substitution $f(x) = g(x)/x$ , which gives us
$\frac{g(y g(x)/x)}{y} + \frac{g(xy)}{x} = \frac{g(x/y)}{x} + \frac{g(y/x)}{y}.$ Multiplying both sides by $x$ and applying the substitution $c = xy$ , $d = x/y$ gives
$d g(b) + g(c) = g(d) + d g(1/d), \tag{1}$ where $b = g(\sqrt{cd})/d$ . In actuality $b$ will not be important at all, we just need to know that such $b$ exists.

First, (1) gives
$g(c) < g(d) + d g(1/d) \quad \forall c, d \in \mathbb{R}^+,$ so $g$ is bounded. Let $M = \sup\{g(x) : x \in \mathbb{R}^+\}$ . Then the above also gives $g(d) \leq M \leq g(d) + dM$ , or $(1 - d)M \leq g(d) \leq M$ , for all $d \in \mathbb{R}^+$ , so we get $\lim_{d \to 0^+} g(d) = M$ . Finally, rearrange (1) to
$(g(c) - M) = (g(d) - M) + d (g(1/d) - g(b))$ for some $b \in \mathbb{R}^+$ , which yields
$|g(c) - M| \leq |g(d) - M| + d |g(1/d) - g(b)| \leq |g(d) - M| + dM.$ Taking $d \to 0^+$ gives $|g(c) - M| = 0$ , and thus $g(c) = M$ for all $c \in \mathbb{R}^+$ . That is, $g$ is constant, and thus $f(x) = f(1)/x$ for all $x \in \mathbb{R}^+$ , which obviously work.

This post has been edited 1 time. Last edited by BlazingMuddy, Apr 19, 2025, 11:38 PM
Reason: Fix formatting

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#8 h Apr 21, 2025, 8:48 PM

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Official (my) solution:

Introduce the function $g(x) = x f(x)$ and rewrite the condition using it:
$g\left(\frac{y}{x} g(x)\right) + \frac{y}{x} g(xy) = \frac{y}{x} g\left(\frac{x}{y}\right) + g\left(\frac{y}{x}\right)$
Substitute $z = \frac{y}{x}$ , and we obtain an identity that holds for all $x, z > 0$ :
$g(zg(x)) + zg(x^2 z) = zg\left(\frac{1}{z}\right) + g(z) \tag{1}$
Substitute $z = 1$ into (1):
$g(g(x)) + g(x^2) = 2g(1) \quad \Rightarrow \quad g(x^2) < 2g(1)$
Since $x^2$ takes on all positive values, we conclude that $g(x) < 2g(1)$ for all $x > 0$ .

Now let $a, b > 0$ be arbitrary numbers. Choose $x, z$ such that $x^2 z = a$ , $\frac{1}{z} = b$ (i.e., $x = \sqrt{ab}$ , $z = \frac{1}{b}$ ), and rewrite the identity:

$g\left( \frac{g(\sqrt{ab})}{b} \right) + \frac{g(a)}{b} = \frac{g(b)}{b} + g\left( \frac{1}{b} \right) \quad \Rightarrow \quad g(a) - g(b) = b \left( g\left( \frac{1}{b} \right) - g\left( \frac{g(\sqrt{ab})}{b} \right) \right)$
The absolute value of the right-hand side does not exceed $b \cdot 2g(1)$ , so for any $a, b > 0$ we have:
$|g(a) - g(b)| \leq 2g(1) b \tag{2}$
Suppose there exist $a_1, a_2$ such that $g(a_1) \ne g(a_2)$ . Then, from (2), for any $b > 0$ we have:
$|g(a_1) - g(a_2)| \leq |g(a_1) - g(b)| + |g(b) - g(a_2)| \leq 4g(1) b$
This is impossible, since $b$ can be chosen arbitrarily small. Hence, $g(x) \equiv k$ — a constant, and then $f(x) = \frac{k}{x}$ .

Substitute into the original condition and verify that this $f$ satisfies it.

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