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Source: Mongolian TST 2025 P1.

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B1t

#1 h Apr 26, 2025, 6:45 AM • 1 Y

Y by farhad.fritl

Denote the set of real numbers by $\mathbb{R}$ . Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that for all $x, y, z \in \mathbb{R}$ ,
$f(xf(x+y)+z) = f(z) + f(x)y + f(xf(x)).$

This post has been edited 2 times. Last edited by B1t, Apr 26, 2025, 7:01 AM

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NicoN9

#2 h Apr 26, 2025, 6:48 AM

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$f(x)f(x)$ means $f(x)^2$ ? or is it a typo?

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#3 h Apr 26, 2025, 6:55 AM

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$x = y = z = 0$ implies $f(0) = 0$
$x = -y$ implies $f(z) = f(z) - xf(x) + f(x)^2,$ so $f\in\{0, x\}.$
assume $f(a ) = 0, f(b ) = b.$
take $y = 0,$ and we get $f(xf(x) + z) = f(z) + f(x)^2.$
if we take $x = b, y = a - b, z = a,$ we get $0 = ab.$ this implies $f \equiv 0, x.$

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B1t

#5 h Apr 26, 2025, 7:05 AM

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NicoN9 wrote:

$f(x)f(x)$ means $f(x)^2$ ? or is it a typo?

I wrote it wrong. im sorry

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#6 h Apr 26, 2025, 7:20 AM

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suppose $f$ is not identically $0$
suppose $x_0$ satisfies $f(x_0)\neq 0,$ and if we pick $x = x_0$ and range the values of $y,$ we obtain that $f$ is surjective. ignore the definition of $x_0$
take $y = 0$ and we get $f(xf(x) + z) = f(xf(x)) + f(z)$
take $x = - y$ yields $f(xf(x))=xf(x).$
thus, $f(f(xf(x)) + z) = f(z) + f(f(xf(x)))$ which implies that the function is cauchy
plugging back into original equation, one gets $f(xf(y)) = f(x)y$
pick $x = 1$ to get $f(f(y)) = y,$ which is well-known to imply that $|f(y)| = |y|.$

assume $A$ is the set of $x$ such that $f(x) = x$ and $B$ is the set of $x$ such that $f(x) = -x.$
if $x, y \in A,$ then $f(xy) = xy,$ so $xy \in A$
if $x\in A, y\in B,$ then $f(xy) = -xy,$ so $xy\in B$
if $x \in B, y \in A,$ then $f(xy) = -xy,$ so $xy \in B$
if $x \in B, y\in B,$ then $f(xy) = xy,$ so $xy\in A$

this implies $f$ is multiplicative
hence, since $f$ is both multiplicative and additive, $f$ is identity

This post has been edited 1 time. Last edited by Ilikeminecraft, Apr 26, 2025, 3:22 PM

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Haris1

#7 h Apr 26, 2025, 8:57 AM

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I wont write the solution , i will just write the steps.
$1.$ Prove that the function is additive
$2.$ Prove that its either constant or bijective
$3.$ Prove that its multiplicative and finish

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552 posts

#8 h Apr 26, 2025, 12:40 PM • 1 Y

Y by farhad.fritl

B1t wrote:

Denote the set of real numbers by $\mathbb{R}$ . Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that for all $x, y, z \in \mathbb{R}$ ,
$P(x,y,z):\, f(xf(x+y)+z) = f(z) + f(x)y + f(xf(x)).$

$P(0,0,0)$ $\implies$ $f(0)=0$ .
$P(x,-x,0)$ $\implies$ $f(xf(x))=xf(x)$
So $P(x,y-x,z): \, f(xf(y)+z)=f(z)+yf(x)$ (new $P$ )
- If $f(t)=0$ for some $t \neq 0$ , then $P(x,t,0)$ $\implies$ $\boxed{f=0}$ which is a solution.
- Suppose $f(t)=0$ $\iff$ $t=0$ .
$P(1,x,0)$ $\implies$ $f(f(x))=xf(1)$ . hence $f$ is bijective, replace $x=1$ in this equation, we get $f(f(1))=f(1)$ so $f(1)=1$ .
Therefore $f(f(x))=x$ for all $x$ .
$P(x, f(y), 0)$ $\implies$ $f(xy)=f(x)f(y)$ for all $x, y$
$P(x, 1, y)$ $\implies$ $f(x+y)=f(x)+f(y)$ for all $x, y$
So $f$ is additive and multiplicative, hence $\boxed{f=\mathrm{Id}}$ which is also a valid solution.

This post has been edited 1 time. Last edited by cazanova19921, Apr 26, 2025, 1:55 PM

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1525 posts

#9 h Apr 27, 2025, 3:25 AM

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Denote $P(x,y,z)$ the assertion of the given F.E.
Notice from $P(0,y,z)$ we can inmediately get $f(0)=0$ and also $P(x,-x,z)$ gives that $f(xf(x))=xf(x)$ but also $P(x,y,0)$ now gives that $f(xf(x+y))=(x+y)f(x)$ so in fact shifting gives $Q(x,y)$ which is $f(xf(y))=yf(x)$ now notice if ther existed some $c \ne 0$ for which $f(c)=0$ then $Q(x,c)$ gives that $f(x)=0$ for all reals $x$ so its either that or $f$ is injective at zero, but now basically note that our F.E. may now by re-written as $R(x,y,z)$ to be $f(xf(y)+z)=f(z)+yf(x)$ for all reals $x,y,z$ but also take $y \ne 0$ and shift $x$ to get that $f$ is additive, but also from $Q(x, f(x))$ we get that $f(f(1)x^2)=f(x)^2$ and therefore shifting $x \to f(x)$ gives $f(f(1)f(x)^2)=f(1)^2 \cdot x^2$ so for example $f$ is surjective on all non-negative reals also from $Q(x,y)$ we have $f$ injective trivially when setting $x \ne 0$ and this take $f(d)=1$ and $Q(x,d)$ to get that $d=1$ and thus $f(1)=1$ so $Q(1,x)$ gives $f$ is an involution so by $Q(x,f(y))$ we get $f$ multiplicative so addivitive+multiplicative means $f$ is the identity function or constant (later case gives $f$ is zero everywhere so doesn't count), so $f(x)=0,x$ for all reals $x$ are the only two solutions that work thus we are done :cool:

:cool:

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This post has been edited 1 time. Last edited by MathLuis, Apr 27, 2025, 3:32 PM

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#10 h Apr 27, 2025, 4:33 AM

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B1t wrote:

Denote the set of real numbers by $\mathbb{R}$ . Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that for all $x, y, z \in \mathbb{R}$ ,
$f(xf(x+y)+z) = f(z) + f(x)y + f(xf(x)).$

Let $P(x,y)$ be the assertion $f(xf(x+y)+z)=f(z)+yf(x)+f(xf(x))$ .
Note that $\boxed{f(x)=0}$ is a solution, else there is some $j$ with $f(j)\ne0$ .
$P(0,0,0)\Rightarrow f(0)=0\Rightarrow j\ne0$
$P(j,-j,0)\Rightarrow f(jf(j))=jf(j)$
$P(j,0,x)\Rightarrow f(x+jf(j))=f(x)+jf(j)$
$P(x,y+jf(j),z)\Rightarrow f(xf(x+y)+jf(j)x+z)=f(z)+yf(x)+f(xf(x))+jf(j)f(x)$
Comparing this last line with $P(x,y)$ we get:
$f(xf(x+y)+jf(j)x+z)=f(xf(x+y)+z)+jf(j)f(x)$ and setting $z=-xf(x+y)$ and $x=1$ this is $jf(j)(f(1)-1)=0$ , so $f(1)=1$ .
$P(1,x-1,y)\Rightarrow f(f(x)+y)=x+f(y)$
Call this assertion $Q(x,y)$ .
$Q(x,0)\Rightarrow f(f(x))=x$
$Q(f(x),y)\Rightarrow f(x+y)=f(x)+f(y)$
$P(x,0,y)\Rightarrow f(xf(x)+y)=f(xf(x))+f(y)$
$P(x,-x,0)\Rightarrow f(xf(x))=xf(x)$
Now $P(x,y)$ becomes:
$\begin{align*} xf(x)+f(z)+f(xf(y))&=f(xf(x))+f(z)+f(xf(y))\\ &=f(xf(x)+xf(y)+z)\\ &=f(xf(x+y)+z)\\ &=f(z)+yf(x)+f(xf(x))\\ &=f(z)+yf(x)+xf(x) \end{align*}$ and so $f(xf(y))=yf(x)$ . Taking $y\mapsto f(y)$ we have $f(xy)=f(x)f(y)$ , well-known that the only additive and multiplicative functions are $\boxed{f(x)=x}$ and $\boxed{f(x)=-x}$ which both work.

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#11 h Apr 27, 2025, 7:26 AM

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why is mongolian tst so easy?

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B1t

#12 h Apr 27, 2025, 1:30 PM

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GreekIdiot wrote:

why is mongolian tst so easy?

true

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#13 h Apr 27, 2025, 2:14 PM • 2 Y

Y by B1t, Nuran2010

Interesting problem!

Let $P(x, -x, y)$ denote the assertion of the functional equation:
$xf(x) = f(xf(x)) \tag{1}$
Now, consider the following:

- $P(1, -1, x)$ gives:
$f(1) = f(f(1)).$ - $P(1, x-1, 0)$ gives:
$f(f(x)) = x \cdot f(1). \tag{2}$
Now, we consider two cases:

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**Case 1:** $f(1) = 0$

From equation (2), we have:
$f(f(x)) = 0 \quad \text{for all } x.$
Applying $f$ to both sides of equation (1):
$xf(x) = f(xf(x)) = f(f(xf(x))) = 0,$ thus implying:
$f(x) = 0 \quad \text{for all } x.$
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**Case 2:** $f(1) \neq 0$

From equation (2), we can conclude that $f$ is bijective.

Suppose there exists some $k$ such that $f(k) = 1$ .
Applying $P(k, -k, x)$ gives:
$k = 1,$ thus $k = 1$ .

Moreover, from (2), we obtain:
$f(f(x)) = x.$
Now, consider $P(1, f(x) - 1, y)$ . We get:
$f(x) + f(y) = f(x+y),$ meaning $f$ is **additive**.

Since $f$ is additive and satisfies $f(f(x)) = x$ , we deduce that $f(x) = cx$ for some constant $c$ . Substituting back into the original functional equation shows that $c = 1$ , and thus:
$f(x) = x.$
---

**Final answer:**
1. $f(x) = 0$ for all real numbers $x$ , or
2. $f(x) = x$ for all real numbers $x$ .

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#14 h Apr 27, 2025, 3:04 PM

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Denote by $P(x, y, z)$ the assertion of the given functional equation. Then $P(0, 0, 0)$ and $P(x, -x, 0)$ gives the following

Claim 1. $f(0) = 0$ and $f(xf(x)) = xf(x)$ .

Now we can rewrite the given functional equation as
$f(xf(x + y) + z) = f(z) + f(x)(x + y) \qquad (\heartsuit)$ and let $P(x, y, z)$ re-denote the assertion of equation $(\heartsuit)$ and set $f(1) = a$ .

Claim 2. If $a = 0$ , then $f(x) \equiv 0$ .

Proof. $P(x, 1 - x, z)$ implies $f(z) = f(ax + z) = f(z) + f(x)$ , so $f(x) = 0$ for all $x \in \mathbb{R}$ .

In what follows, we always assume $a \ne 0$ .

Claim 3. $f(x)$ is bijective when $a \ne 0$ .

Proof. From $P(1, y, 0)$ we get $f(f(1 + y)) = a(y + 1) \implies f(f(y)) = ay$ . This is a bijection whenever $a \ne 0$ .

Claim 4. $a = f(1) = 1$ and $f(x)$ is additive, i.e. $f(x + y) = f(x) + f(y)$ holds.

Proof. Since $xf(x) = f(xf(x))$ , substituting $x = 1$ we obtain $f(1) = f(f(1)) \implies f(1) = 1$ as $f$ is injective. Then $P(x, 1 - x, z)$ gives $f(x + z) = f(z) + f(x)$ .

Claim 5. $f(x)$ is multiplicative, i.e. $f(xy) = f(x)f(y)$ holds.

Proof. We have $f(f(y)) = ay = y$ so $P(x, y, 0)$ gives $f(xf(x + y)) = f(x)(x + y) = f(x)\cdot f(f(x + y))$ . Then note that
$S_x := \{f(x + y): y \in \mathbb{R}\} = \mathbb{R}$ for fixed $x \in \mathbb{R}$ since $f$ is surjective.

It's well-known that if $f(x)$ is both additive and multiplicative, then either $f(x) \equiv 0$ or $f(x) = x$ .

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#15 h Apr 30, 2025, 12:37 PM

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The claims I made in order to solve:
1) $f(0) = 0$ .
2) $f(xf(x)) = xf(x)$ .
3) $f(xf(y)) = yf(x)$ .
4) Either $f \equiv 0$ or $f(x) = 0 \implies x = 0$ .
(Now onwards taking second case...)
5) $f$ is an involution
6) $f$ is additive
7) $f$ is multiplicative
And done!

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