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Source: 2023 Israel TST Test 7 P1

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#1 h May 9, 2023, 6:56 PM • 2 Y

Y by tiendung2006, dxd29070501

Find all functions $f:\mathbb{R}\to \mathbb{R}$ such that for all $x, y\in \mathbb{R}$ the following holds:
$f(x)+f(y)=f(xy)+f(f(x)+f(y))$

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Prod55

#2 h May 9, 2023, 8:04 PM • 3 Y

Y by tiendung2006, Hoanglahoang, ItsBesi

Very nice!

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

Clearly $f\equiv c$ works, so let's now assume that $f$ is non-constant.

$P(0,0): f(0)=f(2f(0))$
$P(x,0): f(x)=f(f(x)+f(0))$
$P(x,2f(0)): f(x)+f(2f(0))=f(2f(0)x)+f(f(x)+f(2f(0))\Leftrightarrow f(2f(0)x)=f(0)$
so $f(0)=0$ otherwise $f$ must be constant, $P(x,0)$ becomes $f(f(x))=f(x)$ .
Observe that $f(a)=f(0)\Leftrightarrow a=0$ since $P(x,a)$ gives $f(ax)=f(0)=0$ and $f$ is not constant.
Now lets define the set $A=\left \{k| f(x)=f(kx)\forall x \in \mathbb{R} \right \}$
If $f(a)=f(b)$ for some $a\neq b$ then from $P(x,a),P(x,b)$ we get that $f(xa)=f(xb)$ and thus $a/b=k\in A$
Also $f(f(1))=f(1)$ so $f(1)\in A$ ( note that $f(1)\neq 0$ since otherwise
$P(x,1): f(1)=f(f(x)+f(1))\Rightarrow f(f(x))=0$ but $f(f(x))=f(x)$ and hence $f\equiv 0$ a contradiction.)
Now $P(1,1): f(1)=f(2f(1))$ so $2f(1)\in A$ as well and hence for every $x$ we have that
$f(2f(1)x)=f(x)=f(xf(1))\Rightarrow f(x)=f(2x)$ for all $x$ , i.e $2\in A$ .
$P(x,x): 2f(x)=f(x^2)+f(2f(x))=f(x^2)+f(f(x))=f(x^2)+f(x)\Leftrightarrow f(x)=f(x^2)\Rightarrow x\in A,\forall x\neq 0$ .
Now pick any $x,y\neq 0$ , then $x/y\in A$ and thus $f(y)=f(y\cdot x/y)=f(x)$ .
So all the solution are:
$f(x)\equiv c,\forall x$
$f(0)=0,f(x)=c\neq 0,\forall x\neq 0$
and they clearly work.

This post has been edited 1 time. Last edited by Prod55, May 9, 2023, 8:05 PM

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#3 h May 9, 2023, 10:47 PM • 1 Y

Y by Math.1234

Prod55 wrote:

Very nice!

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

Clearly $f\equiv c$ works, so let's now assume that $f$ is non-constant.

$P(0,0): f(0)=f(2f(0))$
$P(x,0): f(x)=f(f(x)+f(0))$
$P(x,2f(0)): f(x)+f(2f(0))=f(2f(0)x)+f(f(x)+f(2f(0))\Leftrightarrow f(2f(0)x)=f(0)$
so $f(0)=0$ otherwise $f$ must be constant, $P(x,0)$ becomes $f(f(x))=f(x)$ .
Observe that $f(a)=f(0)\Leftrightarrow a=0$ since $P(x,a)$ gives $f(ax)=f(0)=0$ and $f$ is not constant.
Now lets define the set $A=\left \{k| f(x)=f(kx)\forall x \in \mathbb{R} \right \}$
If $f(a)=f(b)$ for some $a\neq b$ then from $P(x,a),P(x,b)$ we get that $f(xa)=f(xb)$ and thus $a/b=k\in A$
Also $f(f(1))=f(1)$ so $f(1)\in A$ ( note that $f(1)\neq 0$ since otherwise
$P(x,1): f(1)=f(f(x)+f(1))\Rightarrow f(f(x))=0$ but $f(f(x))=f(x)$ and hence $f\equiv 0$ a contradiction.)
Now $P(1,1): f(1)=f(2f(1))$ so $2f(1)\in A$ as well and hence for every $x$ we have that
$f(2f(1)x)=f(x)=f(xf(1))\Rightarrow f(x)=f(2x)$ for all $x$ , i.e $2\in A$ .
$P(x,x): 2f(x)=f(x^2)+f(2f(x))=f(x^2)+f(f(x))=f(x^2)+f(x)\Leftrightarrow f(x)=f(x^2)\Rightarrow x\in A,\forall x\neq 0$ .
Now pick any $x,y\neq 0$ , then $x/y\in A$ and thus $f(y)=f(y\cdot x/y)=f(x)$ .
So all the solution are:
$f(x)\equiv c,\forall x$
$f(0)=0,f(x)=c\neq 0,\forall x\neq 0$
and they clearly work.

l hope that my solution is true
Let $P(x,y)$ be the given assertion
1) $f(0)$ $\neq$ $0$
$P(x,0)$ $\rightarrow$ $f(x)=f(f(x)+f(0))$ $(1)$
$P(x,1)$ $\rightarrow$ $f(f(x)+f(1))=f(1)$ $(2)$
From $(1)$ we get $f(0)=f(2f(0))$
$P(\frac{1}{2f(0)},2f(0))$ $\rightarrow$ $f(0)=f(1)$ and from $(1)$ and $(2)$ we get f(x)=f(1)=c for all $x$ $\in$ $R$ $c$ $\neq$ $0$
2) $f(0)=0$
$P(x,0)$ $\rightarrow$ $f(x)=f(f(x))$ $(1)$
$P(x,1)$ $\rightarrow$ $f(f(x)+f(1))=f(1)$ $(2)$
From $(2)$ we get $f(2f(1)$ $=$ $f(1)$ then $P(2,f(1))$ $\rightarrow$ $f(2)=f(1)$ $(3)$
From $P(f(x),y)$ $=G(x,y)$ and $(1)$ we get $f(f(x)y)=f(xy)$ from $G(1,x)$ , $G(2,x)$ and $(3)$ we get $f(x)=f(2x)$ from here we get
$f(x)=f(\frac{x}{2^{\infty}})=f(0)=0$
Then for all $x$ $\in$ $R$ $f(x)=0$

This post has been edited 1 time. Last edited by Mathlover_1, May 9, 2023, 10:48 PM

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#4 h May 9, 2023, 11:25 PM • 1 Y

Y by bin_sherlo

Neat F.E., let $P(x,y)$ the assertion as usual, assume $f$ non-constant as the other case works.
Main Weapon: If $f(a)=f(b)$ , from $P(x,a)-P(x,b)$ we get $f(ax)=f(bx)$ for all $x \in \mathbb R$ call this $\star$
Now $P(0,0)$ gives $f(0)=f(2f(0))$ so from $\star$ we have $f(2f(0)x)=f(0)$ so if $f(0) \ne 0$ then $f$ is constant so we get that $f(0)=0$ , now clearly $f$ is injective at $0$ or else $f(cx)=0$ for $c \ne 0$ such that $f(c)=0$ which again gives $f$ constant so nope.
Now $P(1,1)$ gives $f(1)=f(2f(1))$ and $P(x,0)$ gives $f(x)=f(f(x))$ so by $\star$ we get $f(2f(1)x)=f(x)=f(f(1)x)$ so $f(x)=f(2x)$ and now by $P(x,x)$ we get that $f(x)=f(x^2)$ so by $\star$ we get that $f(xy)=f(xy^2)$ , now pick $x,y \ne 0$ and let $x=\frac{a}{y}$ and $y=\frac{b}{a}$ for any reals $a,b \ne 0$ and we get $f(a)=f(b)$ so $f$ constant for all values but $0$ .
Hence the set of solutions is $f(x)=c$ for all reals $x$ and $f(x)=c \ne 0$ for all $x \ne 0$ and $f(0)=0$ , thus we are done

.

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Amitai

#5 h May 23, 2023, 7:28 PM

Y by

Let P(x,y) be the assertion: f(x)+f(y)=f(xy)+f(f(x)+f(y))
assume f(a)=f(b). If we substitute y=a or b, the only term that can change is f(ax). Therfore f(ax)=f(bx) for all x. P(0,0): f(2f(0))=f(0) => f(2f(0)x)=f(0). So either f is constant (which works) or f(0)=0.
P(x,0): f(x)=f(f(x)) => f(2x)=f(2f(x))
(for x,y≠0) f(P(x,y)): f(f(x)+f(y))=f(f(xy)+f(f(x)+f(y)))=
=f(f(x)+f(y))+f(xy)-f(xy(f(x)+f(y))) => f(f(x)+f(y))=f(1).
For x=y≠0 we have f(1)=f(2f(x))=f(2x).
So f is constant except 0 which it is 0 at (it works).
But what if f was from R+ to R+, you might ask.

Well: the equation f(f(x)+f(y))=f(1) still holds, so if we can rewrite the equation as f(x)+f(y)=f(xy)+f(1).
Denote m=Inf(Im(f)). Suppose f(x)<m+ε for arbitrarily small ε>0.
P(x,x): f(1)=2f(x)-f(x^2)<m+2ε. so f(1)=m.
P(x,1/x): f(x)+f(1/x)=2m => f(x)=m for all x. So the only solution is a positive constant function.

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PotatoTheMathematician

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PotatoTheMathematician

#6 h May 24, 2023, 1:00 PM

Y by

Mathlover_1 wrote:

Prod55 wrote:

Very nice!

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

Clearly $f\equiv c$ works, so let's now assume that $f$ is non-constant.

$P(0,0): f(0)=f(2f(0))$
$P(x,0): f(x)=f(f(x)+f(0))$
$P(x,2f(0)): f(x)+f(2f(0))=f(2f(0)x)+f(f(x)+f(2f(0))\Leftrightarrow f(2f(0)x)=f(0)$
so $f(0)=0$ otherwise $f$ must be constant, $P(x,0)$ becomes $f(f(x))=f(x)$ .
Observe that $f(a)=f(0)\Leftrightarrow a=0$ since $P(x,a)$ gives $f(ax)=f(0)=0$ and $f$ is not constant.
Now lets define the set $A=\left \{k| f(x)=f(kx)\forall x \in \mathbb{R} \right \}$
If $f(a)=f(b)$ for some $a\neq b$ then from $P(x,a),P(x,b)$ we get that $f(xa)=f(xb)$ and thus $a/b=k\in A$
Also $f(f(1))=f(1)$ so $f(1)\in A$ ( note that $f(1)\neq 0$ since otherwise
$P(x,1): f(1)=f(f(x)+f(1))\Rightarrow f(f(x))=0$ but $f(f(x))=f(x)$ and hence $f\equiv 0$ a contradiction.)
Now $P(1,1): f(1)=f(2f(1))$ so $2f(1)\in A$ as well and hence for every $x$ we have that
$f(2f(1)x)=f(x)=f(xf(1))\Rightarrow f(x)=f(2x)$ for all $x$ , i.e $2\in A$ .
$P(x,x): 2f(x)=f(x^2)+f(2f(x))=f(x^2)+f(f(x))=f(x^2)+f(x)\Leftrightarrow f(x)=f(x^2)\Rightarrow x\in A,\forall x\neq 0$ .
Now pick any $x,y\neq 0$ , then $x/y\in A$ and thus $f(y)=f(y\cdot x/y)=f(x)$ .
So all the solution are:
$f(x)\equiv c,\forall x$
$f(0)=0,f(x)=c\neq 0,\forall x\neq 0$
and they clearly work.

l hope that my solution is true
Let $P(x,y)$ be the given assertion
1) $f(0)$ $\neq$ $0$
$P(x,0)$ $\rightarrow$ $f(x)=f(f(x)+f(0))$ $(1)$
$P(x,1)$ $\rightarrow$ $f(f(x)+f(1))=f(1)$ $(2)$
From $(1)$ we get $f(0)=f(2f(0))$
$P(\frac{1}{2f(0)},2f(0))$ $\rightarrow$ $f(0)=f(1)$ and from $(1)$ and $(2)$ we get f(x)=f(1)=c for all $x$ $\in$ $R$ $c$ $\neq$ $0$
2) $f(0)=0$
$P(x,0)$ $\rightarrow$ $f(x)=f(f(x))$ $(1)$
$P(x,1)$ $\rightarrow$ $f(f(x)+f(1))=f(1)$ $(2)$
From $(2)$ we get $f(2f(1)$ $=$ $f(1)$ then $P(2,f(1))$ $\rightarrow$ $f(2)=f(1)$ $(3)$
From $P(f(x),y)$ $=G(x,y)$ and $(1)$ we get $f(f(x)y)=f(xy)$ from $G(1,x)$ , $G(2,x)$ and $(3)$ we get $f(x)=f(2x)$ from here we get
$f(x)=f(\frac{x}{2^{\infty}})=f(0)=0$
Then for all $x$ $\in$ $R$ $f(x)=0$

the last step is wrong, $f(\frac{x}{2^{\infty}})=f(0)$ works if $f$ is continuous at 0 but it's not always true.

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cursed_tangent1434

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cursed_tangent1434

#8 h Jun 24, 2024, 3:55 PM • 2 Y

Y by dolphinday, Ali_Vafa

Solved with hints. Really nice and instructive problem. I think I learnt a lot from this one. The answers are $f(x) = c$ for all $x \in \mathbb{R}$ for a fixed constant $c \in \mathbb{R}$ and
$f(x) = \begin{cases} 0 & x=0\\ c & x\neq 0 \end{cases}$ for a fixed constant $c \in \mathbb{R}$ . It’s easy to see that these functions satisfy the given conditions. We now show these are the only solutions. Let $P(x,y)$ denote the assertion that $f(x)+f(y)=f(xy)+f(f(x)+f(y))$ for real numbers $x$ and $y$ . We first start off with fixing certain values of $f$ .

Claim : $f(0)=0$ or $f$ is constant.
Proof : Say there exists some $\alpha \neq 0 \in \mathbb{R}$ such that $f(\alpha)=f(0)$ . Then, comparing $P(x,0)$ and $P(x,\alpha)$ yields,
$f(\alpha x)=f(x)+f(\alpha)+f(f(x)+f(\alpha))=f(x)+f(0)-f(f(x)+f(0))=f(0)$ Thus, letting $x= \frac{t}{\alpha}$ implies that $f(t)=f(0)$ for all real numbers $t$ . Thus, if $f$ is non-constant there does not exist such $\alpha \neq 0$ such that $f(x)=f(\alpha)$ .

Now, $P(0,0)$ implies $f(0)=f(2f(0))$ which according to our previous observation requires, $2f(0)=0$ so we have $f(0)=0$ indeed as claimed.

In what proceed, we assume that $f$ is not constant. First of all, note that $P(x,0)$ implies,
$f(x)=f(x)+f(0)=f(0)+f(f(x)+f(0))=f(f(x))$ and $P(x,1)$ implies
$f(1)=f(f(x)+f(1))$ for all real numbers $x$ . This will come in useful later. Note that $P(f(x),2)$ gives,
$f(2f(x)) = f(f(x))+f(2)-f(f(f(x))+f(2))=f(x)+f(2)-f(f(x)+f(2))$ and $P(x,x)$ also gives
$f(2f(x))=2f(x)-f(x^2)$ Thus, setting these two expressions for $f(2f(x))$ equal we have
$f(x^2) + f(2) = f(x)+ f(f(x)+f(2))$ and further, when $x=1$ in the above result,
$f(2)=f(f(1)+f(2))=f(1)$ Thus, $f(2)=f(1)$ . Now, returning to $P(x,2)$ we have,
$f(x)+f(1)=f(x)+f(2)=f(2x)+f(f(x)+f(2)) = f(2x)+f(f(x)+f(1))=f(2x)+f(1)$ which implies that $f(2x)=f(x)$ for all real numbers $x$ . Then, applying this result to $P(x,x)$ gives
$2f(x)=f(x^2)+f(2f(x))=f(x^2)+f(f(x))=f(x^2)+f(x)$ so, $f(x^2)=f(x)$ for all reals $x$ . Now, from this based on our previous observation, it follows that $f(xy)=f(xy^2)$ for all real numbers $x$ and $y$ . Considering $x= \frac{b}{a}$ and $y= \frac{a^2}{b}$ for any two non-zero $a\neq b$ implies that $f$ is indeed constant over the non-zero reals. Thus, all solutions are indeed as claimed.

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#9 h Apr 16, 2025, 8:10 PM

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The only solutions are $f\equiv c$ for some constant $c$ , and also $f(x) = c \forall x \ne 0$ and $f(0) = 0$ . These clearly work. Now we assume that $f$ is not constant. It suffices to show that there is a constant $c$ for which $f(x) = c \forall x \ne 0$ .

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

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

$P(x,0): f(x) = f(f(x) + f(0))$ .

So $P(0,0): f(2f(0)) = f(0)$ .

$P(x, 2f(0)): f(x) + f(0) = f(2x f(0)) + f(f(x) + f(0))$ .

Thus, $f(2xf(0)) = f(0)$ . Since $f$ is nonconstant, $f(0) = 0$ .

$P(x,0)$ now gives $f(f(x)) = f(x)$ .

$P(x,y)$ with $P(x,f(y))$ gives $f(xy) = f(xf(y))$ , so also $f(x) = f(xf(1))$ .

We have $f(f(x) + f(y) - f(xy) + f(1)) = f(1)$ .

$P(x, 2f(1)): f(x) + f(1) = f(2x) + f(f(x) + f(1))$ , so $f(x) = f(2x)$ . In this way, we get $f(x) = f(2f(x))$ as well, so $P(x,x)$ gives that $f(x) = f(x^2)$ . Now comparing $P(x,y)$ and $P(x,y^2)$ gives $f(xy) = f(xy^2)$ . For $y \ne 0$ , setting $x = \frac 1y$ gives $f(y) = f(1)$ , as desired.

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#10 h Apr 17, 2025, 4:49 PM • 1 Y

Y by AlexCenteno2007

Phorphyrion wrote:

Find all functions $f:\mathbb{R}\to \mathbb{R}$ such that for all $x, y\in \mathbb{R}$ the following holds:
$f(x)+f(y)=f(xy)+f(f(x)+f(y))$

Let $P(x,y)$ be the assertion $f(x)+f(y)=f(xy)+f(f(x)+f(y))$ .
Note that all functions of the form $\boxed{f(x)=c}$ , $c\in\mathbb R$ work, otherwise suppose $f$ is nonconstant.
$P(0,0)\Rightarrow f(0)=f(2f(0))$
$P(x,0)\Rightarrow f(x)=f(f(x)+f(0))$
$P(x,2f(0))\Rightarrow f(2xf(0))=f(0)$
Comparing the last two we get $f(0)=0$ since $f$ is nonconstant, so $f(f(x))=f(x)$ . Repeat the same process with $1$ instead:
$P(1,1)\Rightarrow f(1)=f(2f(1))$
$P(x,1)\Rightarrow f(1)=f(f(x)+f(1))$
And now we can compare:
$P(x,f(1))\Rightarrow f(x)=f(xf(1))$
$P(x,2f(1))\Rightarrow f(x)=f(2xf(1))$
If $f(1)=0$ then $f$ is constant, so $f(1)\ne0$ and we get $f(x)=f(2x)$ .
$P(x,x)\Rightarrow f\left(x^2\right)=f(x)$
$P\left(y,x^2\right)\Rightarrow f\left(x^2y\right)=f(xy)$
So setting $y=\frac1x$ , we get $\boxed{f(x)=\begin{cases}0&\text{if }x=0\\c&\text{if }x\ne0\end{cases}}$ where $c\in\mathbb R_{\ne0}$ (since we had $c=f(1)$ here), which satisfies the equation (by casework on whether $x$ or $y$ is $0$ ).

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