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Source: Baltic Way 2010

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1347 posts

WakeUp

#1 h Nov 19, 2010, 6:41 PM • 4 Y

Y by jhu08, Adventure10, Mango247, and 1 other user

Let $\mathbb{R}$ denote the set of real numbers. Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that
$f(x^2)+f(xy)=f(x)f(y)+yf(x)+xf(x+y)$
for all $x,y\in\mathbb{R}$ .

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pco

#2 h Nov 19, 2010, 7:48 PM • 7 Y

Y by Abdollahpour, jhu08, Adventure10, Mango247, and 3 other users

WakeUp wrote:

Let $R$ denote the set of real numbers. Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that
$f(x^2)+f(xy)=f(x)f(y)+yf(x)+xf(x+y)$
for all $x,y\in\mathbb{R}$ .

Let $P(x,y)$ be the assertion $f(x^2)+f(xy)=f(x)f(y)+yf(x)+xf(x+y)$

$P(0,x)$ $\implies$ $f(0)(f(x)+x-2)$

If $f(0)\ne 0$ , this implies $f(x)=2-x$ which indeed is a solution.

Let us from know consider that $f(0)=0$

$P(x,0)$ $\implies$ $f(x^2)=xf(x)$
Then : $P(x,y)$ $\implies$ $xf(x)+f(xy)=f(x)f(y)+yf(x)+xf(x+y)$
Same : $P(y,x)$ $\implies$ $yf(y)+f(xy)=f(x)f((y)+xf(y)+yf(x+y)$
Subtracting implies $(x-y)f(x+y)-f(x)-f(y))=0$

and so $f(x+y)=f(x)+f(y)$ $\forall x\ne y$

Plugging this in $xf(x)+f(xy)=f(x)f(y)+yf(x)+xf(x+y)$ , we get $f(xy)=f(x)f(y)+yf(x)+xf(y)$ $\forall x\ne y$

$\iff$ $f(xy)+xy=(f(x)+x)(f(y)+y)$

Let then $g(x)=f(x)+x$ . We got :
$g(0)=0$
$g(x+y)=g(x)+g(y)$ $\forall x\ne y$
$g(xy)=g(x)g(y)$ $\forall x\ne y$

From the first, we get $g(-x)=-g(x)$ and so $g(2x+(-x))=g(2x)+g(-x)$ and so $g(2x)=2g(x)$ and so $g(x+y)=g(x)+g(y)$ $\forall x,y$
From the second, we get $g(x(x+1))=g(x)g(x+1)=g(x)^2+g(x)$
But also $g(x(x+1))=g(x^2+x)=g(x^2)+g(x)$
And so $g(x^2)=g(x)^2$ and so $g(xy)=g(x)g(y)$ $\forall x,y$

So :
$g(x+y)=g(x)+g(y)$ $\forall x,y$
$g(xy)=g(x)g(y)$ $\forall x,y$
And so, very classical, $g(x)=x$ and $f(x)=0$

Hence the two solutions :
$f(x)=2-x$
$f(x)=0$

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1347 posts

WakeUp

#3 h Nov 19, 2010, 8:53 PM • 3 Y

Y by jhu08, Adventure10, Mango247

pco wrote:

From the second, we get $g(x(x+1))=g(x)g(x+1)=g(x)^2+g(x)$
But also $g(x(x+1))=g(x^2+x)=g(x^2)+g(x)$
And so $g(x^2)=g(x)^2$ and so $g(xy)=g(x)g(y)$ $\forall x,y$

Hi pco, could you please explain this part of the solution? Note also $f(x)=-x$ is a solution.

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#4 h Nov 20, 2010, 4:02 AM • 3 Y

Y by jhu08, Adventure10, Mango247

SO im right?

x=y
2f(x^2)=f(x)^2+x(f(x)+f(2x))
2f(0)=f(0)^2
f(0)=0
or
f(0)=1/2
let x=x, y=0
f(x^2)+f(0)=f(x)f(0)+xf(x)
f(0)=1/2
f(x^2)+1/2=f(x)(1/2+x)
f(x)=ax+b
ax^2+b+1/2=(ax/2+ax^2+b/2+bx)
b/2=b+1/2
b=-1
a=2
f(x)=2x-1 is the solution
for
f(0)=0
f(x)=x

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23515 posts

pco

#5 h Nov 20, 2010, 8:21 AM • 3 Y

Y by jhu08, Adventure10, Mango247

WakeUp wrote:

pco wrote:

From the second, we get $g(x(x+1))=g(x)g(x+1)=g(x)^2+g(x)$
But also $g(x(x+1))=g(x^2+x)=g(x^2)+g(x)$
And so $g(x^2)=g(x)^2$ and so $g(xy)=g(x)g(y)$ $\forall x,y$

Hi pco, could you please explain this part of the solution?

Yes,

, I wrote too quickly !
First we can see that $g(x)=0$ is a solution (and so $f(x)=-x$ is indeed !
If $g(x)$ is not the all zero function, let then $u$ such that $g(u)\ne 0$ . If $u=1$ , choose instead $u=-1$ .
Then the second equation gives us $g(u)(g(1)-1)=0$ and so $g(1)=1$

Then $g(x(x+1))=g(x)g(x+1)$ (using second equation since $x\ne x+1$ ) $=g(x)(g(x)+g(1))=g(x)^2+g(x)$
But $g(x(x+1))=g(x^2+x)=g(x^2)+g(x)$
And so, $g(x^2)=g(x)^2$ and so the second equation $g(xy)=g(x)g(y)$ is also true if $x=y$
...

WakeUp wrote:

Note also $f(x)=-x$ is a solution.

Yes,

$g(x)=0$ is also a solution (I forgot it)

And so :
$f(x)=0$
$f(x)=-x$
$f(x)=2-x$

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#6 h Dec 23, 2013, 8:34 PM • 4 Y

Y by Jerry37284, jhu08, Adventure10, Mango247

This is just about as interesting as a FE can get while still dying to the standard strategies of plugging stuff in, taking cases, and testing. Nevertheless, it was a fun problem.

Reading this won't teach you anything except for what tricks you should have tried

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Onlygodcanjudgeme

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Onlygodcanjudgeme

#7 h Mar 3, 2014, 5:01 AM • 3 Y

Y by jhu08, EHHSW, Adventure10

put $(x,y) = (0,0)$ then we take that 1) $f(0) =0$ 2) $f(0) =2$
1) put $(x,y) = (x,0)$ then we take that $f(x^2) = x \cdot f(x)$ and this is odd function .
put $(x,y) = (x,-x)$ then we take that f(-x) = x , f(x) =0
2)put $(x,y) = (0,x)$ then we take that f(x) = 2-x
so answer is f(x)=0 , f(x) = -x ,f(x) = 2-x

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#8 h Dec 10, 2018, 3:33 AM • 3 Y

Y by jhu08, Adventure10, Mango247

@Onlygodcanjudgeme :I think you got " $\forall x$ : either $f(x)=0$ , either $f(x)=-x$ " and not "either $f(x)=0$ $\forall x$ , either $f(x)=-x$ $\forall x$ "

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#9 h Aug 3, 2020, 8:03 AM • 3 Y

Y by jhu08, Mango247, Mango247

Let $P(x,y)$ be the given assertion, $P(0,0)\implies 2f(0)=f(0)^2\implies f(0)=0,2$ .
If $f(0)=2$ , then $P(0,x)\implies 4=2f(x)+2x\implies f(x)=2-x \ \ \forall x\in\mathbb{R}$ .
So, if $f(0)=0,$ $P(x,0)\implies f(x^2)=xf(x) \ \ \ (1)$
$P(x,-x)\implies f(x^2)+f(-x^2)=f(x)f(-x)-xf(x) \ \ \ (2)$
Using $P(x,y)$ and $P(-x,-y)$ , we can arrive at $f(x)f(y)+yf(x)+xf(x+y)=f(-x)f(-y)-yf(-x)-xf(-x-y),$ letting $y=0$ , we get $f(x)=-f(-x)$ where $x\ne0$ , so $f$ is odd.
From $(2)$ , using the fact that $f$ is odd, $f(x)(f(x)+x)=0\implies f(x)=0,-x.$ Now, assume that there exists $a,b\in \mathbb{R}, a,b\ne 0$ such that $f(a)=0$ and $f(b)=-b$ , using $(1)$ , $f(a^2)=0, f(b^2)=-b^2$ ,
using $P(a,b)$ we get $f(ab)=af(a+b).$ If $f(ab)=-ab$ , $-b=f(a+b).$ If $f(a+b)=0$ , then $b=0$ , a contradiction.
If $f(a+b)=-(a+b)$ , then $a=0$ , a contradiction. Thus, when $f(ab)=0$ , $f(a+b)=0$ as $a=0$ is a contradiction.
Take $P(a,b)$ and $P(b,a)$ , subtracting one from another gives $f(a^2)-f(b^2)=bf(a)+af(a+b)-af(b)-bf(a+b)$ which simplifies to $b^2=ab\implies b(a-b)=0$ and so $a=b$ since $b\ne 0$ but this means $f(b)=f(a)=0=-b$ , a contradiction.
Hence, we have $f(x)\equiv 0, 2-x, -x$ as solutions and plugging them into the equation, we see that they indeed satisfy. $\blacksquare$

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#10 h Aug 8, 2020, 5:39 PM • 2 Y

Y by mkomisarova, jhu08

Solved together with @blastoor
Let $P(x,y)$ denote assertion of given functional equation.

Note that $P(0,0)$ gives us $2f(0) = f(0)^2$ , which means that $f(0)=0$ or $f(0)=2$ . If $f(0) =2$ , then $P(0,x)$ gives us:
$\begin{align*} f(0) + f(0) = f(0)f(x) + xf(0) \\ 4 = 2f(x) + 2x \\ f(x) = 2 -x \end{align*}$ It is easy to check that function $f(x) =2-x$ works.

From now we assume that $f(0) =0$ . Note that $P(x,0)$ gives us:
$f(x^2) = xf(x) \qquad (1)$ In relation $(1)$ replacing $x$ by $-x$ yields:
$f(x^2) = xf(x) = -xf(-x) \implies -f(x) = f(-x)$ Also note that $P(x,-x)$ gives us:
$\begin{align*} f(x^2) + f(-x^2) = f(x)f(-x) -xf(x) \\ f(x^2) -f(x^2) = -f(x)^2 - xf(x) \\ f(x)^2 = -xf(x) \end{align*}$ We conclude that $f(x) = 0$ or $f(x) =-x$ . Now we are left to escape pointwise trap. Assume that there exist nonzero real numbers $a, b$ such that $f(a) =-a$ and $f(b) = 0$ . Note that $f(b^2) = bf(b) = 0$ and that $P(b,a-b)$ gives us:
$\begin{align*} f(b^2) +f(b(a-b))= f(b)f(a-b) + (a-b)f(b) + bf(a) \\ f(b(a-b)) = -ab \end{align*}$ If $f(b(a-b)) = b^2 -ab$ , then $b^2 =0$ , which is contradiction since $b \ne 0$ . On another hand id $f(b(a-b)=0=ab$ , then one of the numbers $a,b$ is zero, which is again contradiction.

We conclude that $f(x)=0$ , $f(x) = 2 -x$ , $f(x) =-x$ are only solutions.

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508669

#11 h Feb 13, 2021, 6:07 PM • 1 Y

Y by jhu08

WakeUp wrote:

Let $\mathbb{R}$ denote the set of real numbers. Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that
$f(x^2)+f(xy)=f(x)f(y)+yf(x)+xf(x+y)$ for all $x,y\in\mathbb{R}$ .

Let $P(x, y)$ be the assertion. We claim that all solutions to the given functional equation are of form $\boxed{f(x) = 2-x}$ for reals $x$ , $\boxed{f(x) = -x}$ for all reals $x$ and $\boxed{f(x) = 0}$ for all reals $x$ . It is not hard to see they work. Now we show that these are the only such functions.

We see that $P(x, 0) \implies f(0)(f(x) + x - 2) = 0$ , so if $f(0) \neq 0$ , then $\boxed{f(x) = 2-x}$ which is indeed a solution.

Otherwise, let $f(0) = 0 \dots (1)$ . Then by $P(x, 0)$ , we yield that $f(x^2) = xf(x) = -xf(-x)$ (by replacing $x$ by $-x$ ) and so $f$ is odd function. Now $P(x, y) - P(y, x)$ along with $f(x^2) = xf(x)$ gives that $(x-y)(f(x+y)-f(x)-f(y)) = 0$ and so if $x \neq y$ , definitely $f(x+y) = f(x) + f(y)$ and so $f$ is additive. Now, we re-arrange few terms in $P(x, y)$ .

$P(x, y) \implies f(x^2)+f(xy)=f(x)f(y)+yf(x)+xf(x+y) \implies xf(x) + yf(x) = f(x)f(y) + yf(x) + xf(x) + xf(y) \implies 0 = f(x)f(y) + xf(y) = f(y)(f(x) + x)$ which means that $f(x) = -x$ or $f(x) = 0$ for all reals $x$ .

Let us say that $A = \{ x \lvert f(x) = -x, x \neq 0 \}$ and $A = \{ x \lvert f(x) = 0, x \neq 0 \}$ . Let $a \in A, b \in B$ . We see that $f(ab) = bf(a) + af(b)$ , and so here in this case, $f(ab) = b \times -a = a \times 0 = 0$ , so either of $a$ or $b$ is $0$ , a contradiction to definition of elements belonging to sets $A$ and $B$ . Hence, $\lvert A \rvert = 0$ or $\lvert B \rvert = 0$ . We see that $f(0) = 0 = -0$ . Therefore, we see that all solutions to the given functional equation are of form $\boxed{f(x) = 2-x}$ for reals $x$ , $\boxed{f(x) = -x}$ for all positive reals $x$ and $\boxed{f(x) = 0}$ for all reals $x$

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#13 h Jul 17, 2021, 6:31 PM • 1 Y

Y by jhu08

Hint

Let $P(x,y)$ be the assertion $f(x^2)+f(xy)=f(x)f(y)+yf(x)+xf(x+y)$
$P(0,0)\Rightarrow 2f(0)=f(0)^2$
If $f(0)=2$ then:
$P(0,x)\Rightarrow\boxed{f(x)=2-x}$ , which works.
Now assume $f(0)=0$ .
$P(x,0)\Rightarrow f(x^2)=xf(x)$
$P(1,x)\Rightarrow f(x+1)=f(x)+f(1)-f(x)f(1)-xf(1)$
We use this recurrence to find $f(1)$ .
$P(1,1)\Rightarrow f(2)=f(1)-f(1)^2$
$P(1,2)\Rightarrow f(3)=f(2)-f(1)-f(2)f(1)=f(1)^3-2f(1)^2$
$P(1,3)\Rightarrow f(4)=f(3)-2f(1)-f(3)f(1)=-f(1)^4+3f(1)^3-2f(1)^2-2f(1)$
But $f(4)=2f(2)=2f(1)-2f(1)^2$ , so we find that $2f(1)-2f(1)^2=-f(1)^4+3f(1)^3-2f(1)^2-2f(1)$ . Solving, we have $f(1)\in\{-1,0,2\}$ .

$\textbf{Case 1: }f(1)=0$
$P(1,x)\Rightarrow f(x)=f(x+1)$
$P\left(x,\frac yx+1\right)-P\left(x,\frac yx\right)\Rightarrow f(x+y)=f(x)+f(y)$ if $x\ne0$ , but since it holds for $x=0$ , $f$ is additive.
By USAMO 2002/4, since $f(x^2)=xf(x)$ and $f$ is additive, we must have $f(x)=xf(1)$ , hence $\boxed{f(x)=0}$ which works.

$\textbf{Case 2: }f(1)=-1$
$P(1,x)\Rightarrow f(x+1)=2f(x)+x-1$
$P(x,1)\Rightarrow f(x^2)+f(x)=xf(x+1)\Rightarrow xf(x)+f(x)=2xf(x)+x^2-x\Rightarrow\boxed{f(x)=-x}$ since $f(1)=-1$ , which works.

$\textbf{Case 3: }f(1)=2$
$P(1,x)\Rightarrow f(x+1)=-f(x)-2x+2$
$P(x,1)\Rightarrow(x-1)f(x)=-x^2+x\Rightarrow f(x)=\begin{cases}-x&\text{if }x\ne1\\2&\text{if }x=1\end{cases}$ which doesn't work.

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#14 h Aug 8, 2021, 4:35 PM • 1 Y

Y by jhu08

Here is my way to solve the first case where $f(0)=0$

$P(x,0)$ gives us $f(x^2) = xf(x), \forall x \quad (1)$ $P(y,x)$ gives us $f(y^2) + f(xy)= f(x)f(y)+xf(y)+yf(x+y), \forall x,y \quad (2)$ By pluging (1) into (2) then subtracting (2) and the original FE, we got $(x-y) (f(x)+f(y)) = (x-y) f(x+y), \forall x,y$ , which implies $f(x) + f(y) = f(x+y), \forall x \neq y \quad (3)$ From $(1)$ , we also have $f(x^2)= -x f(x)$ which leads to $f(x)= f(-x), \forall x \quad (4)$ By $P(x,-x)$ and using $(4)$ , we got $f(x)^2 = -xf(x), \forall x (5)$ Using $(1)$ and $(3)$ , from the origina FE, we have $f(xy) = f(x)f(y)+xf(y) + yf(x), \forall x \neq y$ , which equivalent to $f(x)(f(y)+y) = f(xy) - xf(y), \forall x \neq y \quad (6)$ Case 1: There is a number $k \neq 0$ such that $f(k) = -k$ . In $(6)$ , let $y=k$ , we have $f(kx)=-kx, \forall x \neq k$ . Thus $f(x) = -x, \forall x \neq k^2$ . In the other hand, $f(k^2) = kf(k) = -k^2$ . So, $f(x)=-x, \forall x$ .
Case 2: There is no number $k$ other than $0$ such that $f(k) = -k$ , which means $f(x) \neq -x, \forall x \neq 0$ . With $(5)$ , we implies $f(x) = 0, \forall x \neq 0$ . In the otherhand, $f(0)=0$ , thus $f(x) = 0$ .

P/S

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#15 h Aug 8, 2021, 6:25 PM • 1 Y

Y by jhu08

WakeUp wrote:

Let $\mathbb{R}$ denote the set of real numbers. Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that
$f(x^2)+f(xy)=f(x)f(y)+yf(x)+xf(x+y)$ for all $x,y\in\mathbb{R}$ .

Case 1: $f$ is constant.
We will set $f(x)=c$ where $c$ is some real constant. Plugging this on the F.E. we get:
$2c=c^2+cy+cx \implies c=0 \implies f(x)=0$ Cade 2: $f$ is non-constant.
Let $P(x,y)$ the assertion of the given F.E.
$P(0,0)$
$f(0)^2=2f(0) \implies f(0)=0 \; \text{or} \; f(0)=2$ Case 2.1: $f(0)=2$
$P(0,x)$
$4=2f(x)+2x \implies f(x)=2-x$ Case 2.2: $f(0)=0$
$P(x,0)$
$f(x^2)=xf(x) \implies f \; \text{odd}$ $P(x,-x)$ where $x$ is any non-cero real
$f(x)^2+xf(x)=0 \implies f(x)=-x$ Since $f(0)=0$ we have that $f(x)=-x \; \forall x \in \mathbb R$
Thus the solutions are:

$\boxed{f(x)=0 \; \forall x \in \mathbb R}$

$\boxed{f(x)=2-x \; \forall x \in \mathbb R}$

$\boxed{f(x)=-x \; \forall x \in \mathbb R}$

Thus we are done :blush:

:blush:

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#16 h Aug 8, 2021, 6:58 PM • 1 Y

Y by jhu08

WakeUp wrote:

Let $\mathbb{R}$ denote the set of real numbers. Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that
$f(x^2)+f(xy)=f(x)f(y)+yf(x)+xf(x+y)$ for all $x,y\in\mathbb{R}$ .

A good problem for oddness of a function and how to tackle pointwise traps. Other steps are natural.

Solution

This post has been edited 3 times. Last edited by rama1728, Aug 8, 2021, 7:14 PM
Reason: .

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JustKeepRunning

2958 posts

JustKeepRunning

#17 h Aug 8, 2021, 9:03 PM • 1 Y

Y by jhu08

A nice exercise for pointwise trap!

The answers are $f\equiv 0, 2-x, -x$ . These work.

Denote the assertion by $P(x,y)$ . $P(0,0)$ gives that $f(0)=0,2$ .

Case 1: $f(0)=2$ .

$P(0,y)$ gives that $f(y)=2-y.$

Case 2: $f(0)=0$

$P(x,0)$ gives that $f(x^2)=xf(x),$ so we have that $f$ is odd. Then $P(x,-x)$ gives that $0=f(x)(f(x)+x),$ so $f(x)=0,-x$ . To avoid pointwise trap, suppose that $f(x)=0$ and $f(y)=-y$ for some $x,y\neq 0$ . Then from $P(x,y)$ and $P(y,x)$ and subtracting, we get that $xf(x)-yf(y)=yf(x)-xf(y)+(x-y)f(x+y)$ . Obviously, $x\neq y,$ and simplifying gives that $f(x+y)=f(y)=-y$ . If $y=0,$ we are done, and if $x+y=y,$ then $x=0,$ and we are done as well.

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#18 h Aug 9, 2021, 4:16 AM • 1 Y

Y by jhu08

JustKeepRunning wrote:

Obviously, $x\neq y,$ and simplifying gives that $f(x+y)=f(y)=-y$ . If $y=0,$ we are done, and if $x+y=y,$ then $x=0,$ and we are done as well.

How did you simplify the equation into this?

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#19 h Aug 9, 2021, 5:26 AM • 1 Y

Y by jhu08

They used the properties $f(x)=0$ and $f(y)=-y$ .

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ZETA_in_olympiad

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ZETA_in_olympiad

#20 h Aug 1, 2022, 11:05 AM

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Let $P(x,y)$ denote the assertion. Quickly $P(0,x)$ gives $f(x)\equiv 2-x$ or $f(0)=0.$ The former works, so we explore the latter.

$P(x,0)$ gives $f(x^2)=xf(x).$ And so comparing $P(x,y)$ with $P(y,x)$ shows that $f(x+y)=f(x)+f(y)$ for all $x\neq y.$ To conclude $P(x,-x)$ implies $f(x)\in \{0,-x\}$ but since additive $f\equiv 0$ or $f\equiv -x$ and both satisfy.

This post has been edited 1 time. Last edited by ZETA_in_olympiad, Aug 1, 2022, 11:08 AM

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#22 h Feb 13, 2025, 3:12 AM

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This dies to basically anything.

Setting $x=y=0$ yields $2f(0) = f(0)^2$ . If $f(0) = 2$ , setting $x=0$ yields $f(x) = 2-x$ immediately.

If $f(0) = 0$ , setting $y=0$ yields $f(x^2) = xf(x)$ , implying $f$ is odd. Setting $y=-x$ in the original, $0=f\left(x^2\right) + f\left(-x^2\right) = f(x)f(-x)-xf(x) = -f(x)^2 - xf(x).$ So for each $x$ , either $f(x) = 0$ or $f(x) = -x$ . There are many ways to resolve the pointwise trap, but here is a really stupid way. By setting $y=x$ we get $f(x)^2 + xf(2x) = xf(x)$ , i.e. $xf(2x) = 2xf(x)$ or $f(2x) = f(x)$ . Furthermore, by swapping $x$ and $y$ , we get $(x-y)f(x) - xf(x+y)=(y-x)f(y) - yf(x+y)$ so $f(x+y) = f(x)+f(y)$ for all $x \neq y$ . Combining this with the previous equation yields that $f$ is Cauchy and bounded below on $x \geq 0$ , thus $f$ is linear. We can check that only $f \equiv 0$ works here.

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#23 h Feb 15, 2025, 5:41 AM

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case 1. f is a constant function :
c+c=c^2+cy+xc x=y=1 plug in we get c=0
so f(x)=0

case 2. f is not a constant function :
P(0,y) :2f(0)=f(0)f(y)+yf(0)

2-1. if f(0) is not 0 then y=1 plug in we get f(1)=1
P(1,y): 1+f(y)=f(y)+y+f(y+1) and we know that f(y)=2-y

2-2.f(0)=0 : P(0,0): f(x^2)=xf(x) this imply f is a odd function
P(x,-x) : 0=-(f(x)^2)-xf(x) we get f(x)=-x

so the answer is f(x)=0 or 2-x or -x

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#24 h Apr 30, 2025, 3:11 AM

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$f(x)=0$ works, now assume $f$ is not constant
$P(0,0)$ gives $f(0)=0, 2$

If $f(0)=2$ , $P(0, x)$ gives $f(x)=2-x$ which works
If $f(0)=0$ , $P(x, 0)$ gives $f(x^2)=xf(x)$ so $f$ is odd
then $P(x, -x)$ gives $-f(x)^2-xf(x)=0$ so $f(x)(f(x)-x)=0$ , so $f(x)=-x, 0$ .
If $f(a)=0$ , $f(b)=-b$ , by $P(a, b)$ and $P(b, a)$ we have contradiction.
So $f(x)=-x$ or $f(x)=0$ in this case.

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