Functional equation from R to R-[INMO 2011]

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Functional equation from R to R-[INMO 2011]

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Potla

#1 h Feb 6, 2011, 12:32 PM • 11 Y

Y by Aryan-23, donotoven, ImSh95, tiendung2006, Adventure10, Mango247, and 5 other users

Find all functions $f:\mathbb{R}\to \mathbb R$ satisfying
$f(x+y)f(x-y)=\left(f(x)+f(y)\right)^2-4x^2f(y),$ For all $x,y\in\mathbb R$ .

This post has been edited 2 times. Last edited by Amir Hossein, May 4, 2018, 3:37 AM

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SCP

#2 h Feb 6, 2011, 1:11 PM • 5 Y

Y by donotoven, ImSh95, Adventure10, Mango247, and 1 other user

Potla wrote:

Let $f:\mathbb{R}\to \mathbb R$ be a functional equation satisfying
$f(x+y)f(x-y)=\left(f(x)+f(y)\right)^2-4x^2f(y),$
For all $x,y\in\mathbb R,$ where $\mathbb R$ denotes the set of all real numbers.

$0,0$ gives $f(0)=0$
$0,y$ gives $f(y)f(-y)=f^2(y)$ and after $2$ cases, there holds always $f(y)=f(-y)$
Then $y,-y$ gives $0=f(0)=4f^2(y)-4y^2f(y)$
and so if $y \not 0$ we see $f(y)=0$ or $f(y)=y^2$
easy to check these are only solutions.

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pco

#3 h Feb 6, 2011, 1:14 PM • 13 Y

Y by Aranya, abmax777, dgmath97, thewitness, lneis1, donotoven, ImSh95, Adventure10, Mango247, and 4 other users

Potla wrote:

Let $f:\mathbb{R}\to \mathbb R$ be a functional equation satisfying
$f(x+y)f(x-y)=\left(f(x)+f(y)\right)^2-4x^2f(y),$
For all $x,y\in\mathbb R,$ where $\mathbb R$ denotes the set of all real numbers.

Let $P(x,y)$ be the assertion $f(x+y)f(x-y)=(f(x)+f(y))^2-4x^2f(y)$

$P(0,0)$ $\implies$ $f(0)^2=4f(0)^2$ $\implies$ $f(0)=0$
$P(x,x)$ $\implies$ $f(x)(f(x)-x^2)=0$ and so $\forall x$ : either $f(x)=0$ , either $f(x)=x^2$

Suppose now that $\exists a\ne 0, b\ne 0$ such that $f(a)=a^2$ and $f(b)=0$
$P(a,b)$ $\implies$ $f(a+b)f(a-b)=a^4\ne 0$
So $f(a+b)\ne 0$ and so $f(a+b)=(a+b)^2$
Same, $f(a-b)\ne 0$ and so $f(a-b)=(a-b)^2$
And our equality becomes $(a^2-b^2)^2=a^4$ and so, since $b\ne 0$ : $b^2=2a^2$
And this is obviously impossible, else :
Choose $c\notin\{0,b,-b,a,-a\}$ :
If $f(c)=0$ , we get (using $a,c$ instead of $a,b$ in lines above : $c^2=2a^2=b^2$ , impossible
If $f(c)=c^2$ , we get, (using $c,b$ instead pf $a,b$ in lines above : $b^2=2c^2$ and so $c^2=a^2$ , impossible.

And so :
either $f(x)=0$ $\forall x$ which indeed is a solution
either $f(x)=x^2$ $\forall x$ which indeed is a soluton.

@SCP : you got " $\forall x$ : either $f(x)=0$ , either $f(x)=x^2$ " and not "either $f(x)=0$ $\forall x$ , either $f(x)=x^2$ $\forall x$ " and you miss the second part of the proof.

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#4 h Feb 6, 2011, 2:04 PM • 4 Y

Y by donotoven, ImSh95, Adventure10, Mango247

Potla wrote:

Let $f:\mathbb{R}\to \mathbb R$ be a functional equation satisfying
$f(x+y)f(x-y)=\left(f(x)+f(y)\right)^2-4x^2f(y),$
For all $x,y\in\mathbb R,$ where $\mathbb R$ denotes the set of all real numbers.

$y=0\Longrightarrow f(0)=0$ or $f(x)=\frac{4x^2-f(0)}{2}$ . Suppose $f(x)=\frac{4x^2-f(0)}{2}$ . $x=0$ implies that $f(0)=0\Longrightarrow f(x)=2x^2$ which when substituted back proves that it is not a solution. Now, $f(0)=0$ . $x=0\Longrightarrow f(x)=0$ for all $x$ or $f$ is even. If $f$ is even, $x=-y$ implies that $f(x)=0$ for all $x$ or $f(x)=x^2$ which are solutions.

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pco

#5 h Feb 6, 2011, 2:11 PM • 4 Y

Y by ImSh95, Adventure10, Mango247, and 1 other user

Goutham wrote:

... $x=-y$ implies that $f(x)=0$ for all $x$ or $f(x)=x^2$ which are solutions.

Wrong : $x=-y$ implies that for all $x$ : $f(x)=0$ or $f(x)=x^2$ , which is quite different from what you wrote.

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#6 h Feb 6, 2011, 2:27 PM • 7 Y

Y by abmax777, CancerPatient, ImSh95, Adventure10, Mango247, and 2 other users

pco wrote:

Suppose now that $\exists a\ne 0, b\ne 0$ such that $f(a)=a^2$ and $f(b)=0$
$P(a,b)$ $\implies$ $f(a+b)f(a-b)=a^4$

We can continue in a much simpler way. We also have $P(b,a)$ $\implies$ $f(b+a)f(b-a)=a^4 - 4b^2a^2$ .

But $f$ is an even function (*) in both cases, so $a^4 = a^4 - 4b^2a^2$ , whence $ab = 0$ , absurd.

(*) Take $b=0$ in the above relation(s) (we can do this, since $f(0) = 0$ ); so $f(a) = a^2$ , but from the second, $f(a)f(-a) = a^4$ , therefore also $f(-a) = a^2$ . This means that the set $A$ of the numbers $a$ with $f(a) = a^2$ is such that $A = -A$ , so also the set $B$ of the numbers $b$ with $f(b) = 0$ is such that $B = -B$ , and my assertion is true.
Even simpler, for $x=0$ (and knowing $f(0)=0$ ), from the initial equation we get $f(y)f(-y) = f(y)^2$ , so $f$ is even.

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#7 h Feb 6, 2011, 3:38 PM • 3 Y

Y by ImSh95, Adventure10, Mango247

[CONTENT DELETED]

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#8 h Feb 6, 2011, 6:09 PM • 3 Y

Y by ImSh95, Adventure10, Mango247

I would like to introduce the correction in the question posted.
Find all functions $f:\mathbb{R}\to \mathbb R$ be a functional equation satisfying
$f(x+y)f(x-y)=\left(f(x)+f(y)\right)^2-4x^2f(y),$
For all $x,y\in\mathbb R,$ where $\mathbb R$ denotes the set of all real numbers.
[moderator edit: thank you, edited.

]
I did something like SCP but i got
$f(x)f(x)= x^2f(x)$ . where x for all x $\in$ reals. as i wrote it in the examination.
now we have two cases f(x) = 0 forms a function satisgying the given functional equation and f(x) = x^2 forms another function.
One of the biggest statement which can cause wrong conclusion seems to be from $f(x)f(x)= x^2f(x)$ .
As per respected 'pco' i forgot to prove the second part as well.

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RSM

#9 h Feb 8, 2011, 11:14 AM • 2 Y

Y by ImSh95, Adventure10

I want to add one more condition with the functional equation problem that x is not equal to y.
Here is its solution:-

Denote the equation $f(x+y)f(x-y)=(f(x)+f(y))^2-4x^2f(y)$ as $(1)$ .
Exchanging x and y, this leads to
$f(x+y)f(y-x)=(f(x)+f(y))^2-4y^2f(x) \ \ \ \ \ (2)$

Putting y=0 in (1), we have
$f(x)^2=(f(x)+c)^2-4cx^2 \ \ \text{where} \ \ c=f(0); \ \ \ \ (3)$

Or,
$f(x)f(-x)=(f(x)+c)^2 \ \ \ \ (4)$

Subtracting $(3)$ from $(4)$ , we have
$f(x)\{f(-x)-f(x)\}=4cx^2 \ \ \ \ \ (5);$

And subtracting $(1)$ from $(2)$ , we have
$f(x+y)[f(x-y)-f(y-x)]=4y^2f(x)-4x^2f(y).$

Applying $(5)$ here and putting $y=0$ and after simplifying this we will get $c=0$ or f is odd.

Case I: c=0.
In this case, by (4) we get f(x)=0 or f is even (this means f is even for all x).

Comparing (1) and (2) we have $\frac{f(a)}{a^2}=\frac{f(b)}{b^2}.$

Since f is even, this holds for every two elements of $\mathbb R$ except $0$ .

So $f(x)=mx^2 \forall x\in\mathbb R.$

Substituting this in (1), we get m=0 or 1.

Case II: f is odd.
Substituting this in $(5)$ we obtain

$f(x)=nx$ for some constant n

Now, it is easy to check that this is inconsistent with the given conditions.

Thanks Potla for Latexifying this.

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#10 h Feb 14, 2011, 6:34 AM • 3 Y

Y by ImSh95, Adventure10, Mango247

After getting $f(x)=0$ or $f(x)=x^2$ and omitting the solution $f(x)=0$ for all real $x$ .It suffices to prove that $f(x)\neq 0$ if $x\neq 0$ .Let $y$ be a non-zero real such that $f(y)=0$ and $x$ be a non-zero real such that $f(x)=x^2$ .
These imply, $f(x+y)f(x-y)=(f(x))^2>0$ (*).So both factors of the RHS are non-zero.so $f(x\pm y)=(x\pm y)^2$ .Substituting this,
(*) reduces to $y^2=2x^2$ .But this makes the number of $y$ 's and $x$ 's finite.Contradiction.

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#11 h Feb 15, 2011, 4:41 AM • 2 Y

Y by ImSh95, Adventure10

Goutham wrote:

Potla wrote:

Let $f:\mathbb{R}\to \mathbb R$ be a functional equation satisfying
$f(x+y)f(x-y)=\left(f(x)+f(y)\right)^2-4x^2f(y),$
For all $x,y\in\mathbb R,$ where $\mathbb R$ denotes the set of all real numbers.

$y=0\Longrightarrow f(0)=0$ or $f(x)=\frac{4x^2-f(0)}{2}$ . Suppose $f(x)=\frac{4x^2-f(0)}{2}$ . $x=0$ implies that $f(0)=0\Longrightarrow f(x)=2x^2$ which when substituted back proves that it is not a solution.(I forgot to substitute it back in the examination

) Now, $f(0)=0$ . $x=0\Longrightarrow f(x)=0$ for all $x$ or $f$ is even. If $f$ is even, $x=-y$ implies that $f(x)=0$ for all $x$ or $f(x)=x^2$ which are solutions.

Sorry Goutham.
Sai Mali, Anshul and me discussed the problem and found f(x)=x^2 and not 0.

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SCP

#12 h Feb 15, 2011, 7:09 PM • 3 Y

Y by ImSh95, Adventure10, Mango247

jgthegreat wrote:

Goutham wrote:

Potla wrote:

Let $f:\mathbb{R}\to \mathbb R$ be a functional equation satisfying
$f(x+y)f(x-y)=\left(f(x)+f(y)\right)^2-4x^2f(y),$
For all $x,y\in\mathbb R,$ where $\mathbb R$ denotes the set of all real numbers.

$y=0\Longrightarrow f(0)=0$ or $f(x)=\frac{4x^2-f(0)}{2}$ . Suppose $f(x)=\frac{4x^2-f(0)}{2}$ . $x=0$ implies that $f(0)=0\Longrightarrow f(x)=2x^2$ which when substituted back proves that it is not a solution.(I forgot to substitute it back in the examination

) Now, $f(0)=0$ . $x=0\Longrightarrow f(x)=0$ for all $x$ or $f$ is even. If $f$ is even, $x=-y$ implies that $f(x)=0$ for all $x$ or $f(x)=x^2$ which are solutions.

Sorry Goutham.
Sai Mali, Anshul and me discussed the problem and found f(x)=x^2 and not 0.

$f=0$ is of course a solution also.

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#13 h Feb 16, 2011, 2:37 AM • 2 Y

Y by ImSh95, Adventure10

jgthegreat wrote:

Sorry Goutham.
Sai Mali, Anshul and me discussed the problem and found f(x)=x^2 and not 0.

Maybe you missed something. Because it is obvious from the first glance that $f(x)=0$ for all $x$ is a solution.

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#14 h Feb 16, 2011, 4:46 AM • 3 Y

Y by ImSh95, Adventure10, Mango247

Let put x=y to get
$f(2x)f(0)=4f(x)^2-4x^2f(x)$
now set x=0 to see
$f(0)^2=4f(0)^2$
so, $f(0)=0$

so put it back to get

$0=4f(x)^2-4x^2f(x)$
$f(x)[f(x)-x^2]=0$

so, $f(x)=0$ or $x^2$

now let there exist $x_{1}\neq 0, x_{2}\neq 0$ s.t. $f(x_{1})=0,f(x_{2})\neq 0$
i.e. $f(x_{1})=0,f(x_{2})=x_{2}^2$

then set $y=x_{1},x=x_{2}$ to get

$f(x_{2}+x_{1})f(x_{2}-x_{1})=x_{2}^4$

since $x_{2}\neq 0$ so, $f(x_{2}+x_{1})\neq 0, f(x_{2}-x_{1})\neq 0$

so, $f(x_{2}+x_{1})=(x_{2}+x_{1})^2, f(x_{2}-x_{1})=(x_{2}-x_{1})^2$

so, $(x_{2}^2-x_{1}^2)^2=x_{2}^4$

or, $x_{1}^2(x_{1}^2+2x_{2}^2)=0$

contradiction as $x_{1}\neq 0, x_{2}\neq 0$ .

so, either $f(x)=0$ for all $x\neq 0$ or $f(x)\neq0$ for all $x\neq 0$ i.e.

$f(x)=0$ for $x\neq 0$ or $f(x)=x^2$ for $x\neq 0$ .

Now see this are indeed solutions.

so only possible solutions are

$f(x)=x^2$ and $f(x)=0$

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CDP100

#15 h May 5, 2011, 5:51 PM • 3 Y

Y by ImSh95, Adventure10, Mango247

sumanguha wrote:

or, $x_{1}^2(x_{1}^2+2x_{2}^2)=0$

I think it must be $x_{1}^2(x_{1}^2-2x_{2}^2)=0$ !

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#16 h May 8, 2011, 3:16 PM • 3 Y

Y by ImSh95, Adventure10, Mango247

Quote:

I think it must be $x_{1}^{2}(x_{1}^{2}-2x_{2}^{2})=0$ !

Yes, you are right.

and then a modified arguement would be.

let there is $x_{1}\neq 0, x_{2}\neq 0$ s.t. $f(x_{1})=0,f(x_{2})\neq 0$

Then by corrected arguement $x_{1}=\pm\sqrt{2} x_{2}$

Then we get either infinitely many (1) $z_{i} \neq 0$ s.t. $f_(z_{i})=0$

or infinitely many (2) $z_{i} \neq 0$ s.t. $f_(z_{i}) \neq 0$

if (1) is true then $x_{2}=\pm\frac{1}{\sqrt{2}}z_{i}$ for all $i$

contradiction since $z_{i}$ are distinct.

if (2) is true then $x_{1}=\pm\sqrt{2}z_{i}$ for all $i$

contradiction since $z_{i}$ are distinct.

so, contradiction.

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Onlygodcanjudgeme

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Onlygodcanjudgeme

#17 h Feb 4, 2014, 2:33 AM • 3 Y

Y by ImSh95, Adventure10, Mango247

$f(x+y) \cdot f(x-y) = [f(x) + f(y)]^2 - 4 \cdot x^2 \cdot f(y)$
$(x,y) = (0,0)$ then we take that $f(0) =0$
$(x.y) = (x,x)$ then we take that $f(x) =0$ or $f(x) = x^2$

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pco

#18 h Feb 4, 2014, 7:52 AM • 3 Y

Y by galav, ImSh95, Adventure10

Onlygodcanjudgeme wrote:

...
$(x.y) = (x,x)$ then we take that $f(x) =0$ or $f(x) = x^2$

No, you should read the entire thread, before posting exactly the same error than above :

$(x.y) = (x,x)$ implies $f(x)(f(x)-x^2)=0$ and so : " $\forall x$ , either $f(x)=0$ , either $f(x)=x^2$ "

Which is quite different from your own conclusion ("either $f(x)=0$ $\forall x$ , either $f(x)=x^2$ $\forall x$ ")
For example the fonction $f(x)=\left(\frac{x+|x|}2\right)^2$ is such that $f(x)(f(x)-x^2)=0$ $\forall x$

So you miss some work again before conclusion.

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Kezer

#19 h Jun 1, 2016, 11:27 AM • 3 Y

Y by ImSh95, Adventure10, Mango247

Just skimmed the thread, seems like my solution hasn't been posted yet.

Set $x=y=0$ , that yields $f(0)=0$ . Now set $x=0$ , that gives $f(y) f(-y) = f(y)^2 \qquad \iff \qquad f(y) \left(f(y)-f(-y) \right) = 0.$ So $f(y)=0$ or $f(y)=f(-y)$ . Let $r \in \mathbb{R}$ be some number for which $f(r)=0$ , assuming such a number exists, for $f(-r)$ we can either have $f(-r)=0$ or $f(-r)=f(r)=0$ , in both cases $f(r)=f(-r)$ . So for each $x \in \mathbb{R}$ we have $f(x)=f(-x)$ . If such a $r$ as above doesn't exist, the conclusion is immediate.
Now exploit the symmetry. As $f(x-y)=f(y-x)$ with $f(x)=f(-x)$ we can see $\left(f(x)+f(y) \right)^2-4x^2f(y) = f(x+y)f(x-y)=f(y+x)f(y-x) = \left(f(y)+f(x) \right)^2 - 4y^2f(x).$ Thus, we have $4x^2f(y) = 4y^2f(x) \qquad \iff \qquad x^2f(y) = y^2f(x).$ Set $y=1$ to get $f(x) = x^2 f(1)$ , hence $f(x)=cx^2$ for some constant $c \in \mathbb{R}$ . Putting that into the equation yields $c=0$ or $c=1$ and $f(x)=0$ and $f(x)=x^2$ are indeed solutions.

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#20 h Dec 25, 2019, 3:45 PM • 2 Y

Y by ImSh95, Adventure10

Where can I find more problems to tackle with Point-wise trap like in this one?

This post has been edited 1 time. Last edited by Math-wiz, Dec 25, 2019, 3:45 PM

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#21 h Dec 25, 2019, 3:49 PM • 2 Y

Y by ImSh95, Adventure10

Math-wiz wrote:

Where can I find more problems to tackle with Point-wise trap like in this one?

Just search up, the following on AoPS search:

+"pointwise"

You'll find tons of problems dealing with point-wise trap...

This post has been edited 2 times. Last edited by AlastorMoody, Dec 25, 2019, 3:53 PM

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#23 h Jan 4, 2020, 5:19 PM • 4 Y

Y by AlastorMoody, ImSh95, Adventure10, Mango247

I found a much simpler solution and so it might be wrong

. Please check it

Let $P(x,y)$ be the assertion. Then $P(0,0)\implies f(0)=0$ and so, $P(0,y)\implies f(y)f(-y)=f(y)^2\implies f(y)=0$ or $f(y)=f(-y)$ . But if $f(y)=0$ then either $f(-y)=0=f(y)$ or $f(y)=f(-y)$ . Thus, $f$ is always an even function. Now, $P(x,y), P(y,x)\implies f(x)/x^2$ is constant for all real $x$ and so $f(x)=cx^2$ for some constant $c$ . From parent equation $c=0,1$ . Thus, $f\equiv 0, f(x)=x^2$ are the only two possible solutions.

This post has been edited 1 time. Last edited by Jupiter_is_BIG, Jan 4, 2020, 6:03 PM

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#25 h Jan 4, 2020, 5:25 PM • 2 Y

Y by ImSh95, Adventure10

Jupiter_is_BIG wrote:

$f(y)f(-y)=f(y)^2\implies f(y)=0$ or $f(y)=f(-y)$ . But if $f(y)=0$ then either $f(-y)=0=f(y)$ or $f(y)f(-y)$

I feel like there's some mistake here.

This post has been edited 1 time. Last edited by Math-wiz, Jan 4, 2020, 5:25 PM

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#26 h Jan 4, 2020, 5:36 PM • 3 Y

Y by ImSh95, Adventure10, Mango247

Potla wrote:

Find all functions $f:\mathbb{R}\to \mathbb R$ satisfying
$f(x+y)f(x-y)=\left(f(x)+f(y)\right)^2-4x^2f(y),$ For all $x,y\in\mathbb R$ .

My solution

$P(0,0)\implies f(0)=0$
$P(0,y)\implies f(y)=f(-y)$
$f(x-y)=f(y-x)$

$f(x+y)f(x-y)=f(x+y)f(y-x)=(f(x)+f(y))^2-4x^2f(y)=(f(x)+f(y))^2-4y^2f(x)$

$x^2f(y)=y^2f(x)$

$f(x)=x^2,0$

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#27 h Jan 4, 2020, 6:04 PM • 2 Y

Y by ImSh95, Adventure10

Math-wiz wrote:

Jupiter_is_BIG wrote:

$f(y)f(-y)=f(y)^2\implies f(y)=0$ or $f(y)=f(-y)$ . But if $f(y)=0$ then either $f(-y)=0=f(y)$ or $f(y)f(-y)$

I feel like there's some mistake here.

What exactly??

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#28 h Jan 5, 2020, 4:31 AM • 2 Y

Y by ImSh95, Adventure10

Jupiter_is_BIG wrote:

Math-wiz wrote:

Jupiter_is_BIG wrote:

$f(y)f(-y)=f(y)^2\implies f(y)=0$ or $f(y)=f(-y)$ . But if $f(y)=0$ then either $f(-y)=0=f(y)$ or $f(y)f(-y)$

I feel like there's some mistake here.

What exactly??

$f(y)(f(y)-f(-y))=0$ . This looks like pointwise trap

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#29 h Jan 5, 2020, 5:18 AM • 6 Y

Y by AlastorMoody, Jupiter_is_BIG, amar_04, ImSh95, Adventure10, Mango247

Jupiter_is_BIG wrote:

I found a much simpler solution and so it might be wrong

. Please check it

Let $P(x,y)$ be the assertion. Then $P(0,0)\implies f(0)=0$ and so, $P(0,y)\implies f(y)f(-y)=f(y)^2\implies f(y)=0$ or $f(y)=f(-y)$ . But if $f(y)=0$ then either $f(-y)=0=f(y)$ or $f(y)=f(-y)$ . Thus, $f$ is always an even function. Now, $P(x,y), P(y,x)\implies f(x)/x^2$ is constant for all real $x$ and so $f(x)=cx^2$ for some constant $c$ . From parent equation $c=0,1$ . Thus, $f\equiv 0, f(x)=x^2$ are the only two possible solutions.

Your solution is correct. In fact the solution in post #19 (2 posts above you) is the same. It's just written in a more detailed way.

This post has been edited 1 time. Last edited by math_pi_rate, Jan 5, 2020, 5:18 AM
Reason: Typo

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#30 h Jan 5, 2020, 5:32 AM • 4 Y

Y by geometrer_enthu, ImSh95, Adventure10, Mango247

@above Thanks for the info

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#31 h Jul 10, 2021, 4:32 AM • 1 Y

Y by ImSh95

math_pi_rate wrote:

Jupiter_is_BIG wrote:

I found a much simpler solution and so it might be wrong

. Please check it

Let $P(x,y)$ be the assertion. Then $P(0,0)\implies f(0)=0$ and so, $P(0,y)\implies f(y)f(-y)=f(y)^2\implies f(y)=0$ or $f(y)=f(-y)$ . But if $f(y)=0$ then either $f(-y)=0=f(y)$ or $f(y)=f(-y)$ . Thus, $f$ is always an even function. Now, $P(x,y), P(y,x)\implies f(x)/x^2$ is constant for all real $x$ and so $f(x)=cx^2$ for some constant $c$ . From parent equation $c=0,1$ . Thus, $f\equiv 0, f(x)=x^2$ are the only two possible solutions.

Your solution is correct. In fact the solution in post #19 (2 posts above you) is the same. It's just written in a more detailed way.

PLEASE read the thread before posting the same error again and again.you have to prove it that both can't be true at a time

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#32 h Jul 10, 2021, 4:52 AM • 1 Y

Y by ImSh95

Let $P(x,y)$ be the assertion $f(x+y)f(x-y)=(f(x)+f(y))^2-4x^2f(y)$ .
$P(0,0)\Rightarrow f(0)=0$
$P(x,x)\Rightarrow f(x)^2=x^2f(x)$
Assume $\exists a,b\ne0:f(a)=0,f(b)=b^2$ .
$\begin{align*} P(b,a)&\Rightarrow f(a+b)f(b-a)=b^4\\ &\Rightarrow f(a+b)^2f(b-a)^2=b^8\\ &\Rightarrow (a+b)^2(b-a)^2f(a+b)f(b-a)=b^8\\ &\Rightarrow (a+b)^2(b-a)^2=b^4\\ &\Rightarrow a^2=2b^2\end{align*}$ For any $j\notin\{0,\pm a,\pm b\}$ , we will have either $f(j)=0$ , in which case $j^2=2b^2=a^2$ , or $f(j)=j^2$ , which would gives $2b^2=a^2=2j^2$ . These are both absurd, hence either $\boxed{f(x)=0}$ or $\boxed{f(x)=x^2}$ which both work.

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#33 h Jul 10, 2021, 5:02 AM • 1 Y

Y by ImSh95

Let $P(x,y)$ denote the given assertion.
$P(0,0) \implies f(0)=0$ .
$P(x,x) \implies f(x)^2 = x^2f(x) \implies f(x)[f(x) - x^2] = 0$ . Thus either $f(x)=0$ for all $x$ or $f(x)=x^2$ for all $x$ .
Now for the pointwise trap, assume there exists $a,b \neq 0$ such that $f(a)=0$ and $f(b)=b^2$ .
$P(b,a) \implies f(b+a)f(b-a) = (b^2)^2 = b^4$ .

Case 1: $f(b+a)=(b+a)^2$ and $f(b-a) = 0$ .
This gives $b=0$ which is a contradiction.

Case 2: $f(b+a) = (b+a)^2$ and $f(b-a)=(b-a)^2$ .
This gives $(b^2 - a^2)^2 = b^4 \implies a^4 - 2a^2b^2 = 0 \implies a=0$ which is a contradiction.

Case 3: $f(b+a)=0$ and $f(b-a)=(b-a)^2$ .
This gives $b=0$ which is a contradiction.

Case 4: $f(b+a)=0$ and $f(b-a)=0$ .
This gives $b=0$ which is a contradiction.

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#34 h Jul 17, 2021, 5:44 AM • 1 Y

Y by ImSh95

MrOreoJuice wrote:

Let $P(x,y)$ denote the given assertion.
$P(0,0) \implies f(0)=0$ .
$P(x,x) \implies f(x)^2 = x^2f(x) \implies f(x)[f(x) - x^2] = 0$ . Thus either $f(x)=0$ for all $x$ or $f(x)=x^2$ for all $x$ .
Now for the pointwise trap, assume there exists $a,b \neq 0$ such that $f(a)=0$ and $f(b)=b^2$ .
$P(b,a) \implies f(b+a)f(b-a) = (b^2)^2 = b^4$ .

Case 1: $f(b+a)=(b+a)^2$ and $f(b-a) = 0$ .
This gives $b=0$ which is a contradiction.

Case 2: $f(b+a) = (b+a)^2$ and $f(b-a)=(b-a)^2$ .
This gives $(b^2 - a^2)^2 = b^4 \implies a^4 - 2a^2b^2 = 0 \implies a=0$ which is a contradiction.

Case 3: $f(b+a)=0$ and $f(b-a)=(b-a)^2$ .
This gives $b=0$ which is a contradiction.

Case 4: $f(b+a)=0$ and $f(b-a)=0$ .
This gives $b=0$ which is a contradiction.

Correct,you have noticed the trap,congrats :cool:

:cool:

But what about case 2,there will be another condition,a=√2b,and the other cases can be ignored as any of them being 0 makes b=0

This post has been edited 1 time. Last edited by HoRI_DA_GRe8, Jul 17, 2021, 5:49 AM

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jelena_ivanchic

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#36 h Jan 6, 2022, 5:36 AM • 1 Y

Y by ImSh95

I did it in way different way and it avoided point wise.
solution

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ZETA_in_olympiad

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#37 h Jun 6, 2022, 6:01 PM • 2 Y

Y by ImSh95, Mango247

Denote the assertion by $P(x,y).$

$P(0,0)$ gives $f(0)=0.$ Moreover $P(0,x)$ yields $f(x)=0$ or $f(x)=f(-x).$ Note that both cases imply $f$ is even. Comparing $P(1,x)$ with $P(x,1)$ we get $f(x)=x^2f(1).$

Checking gives $f(1)\in \{0,1\}$ which correspond to $f\equiv 0$ and $f\equiv \text{Id}^2.$ Clearly both satisfy.

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SomeonecoolLovesMaths

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SomeonecoolLovesMaths

#38 h Sep 27, 2024, 2:19 PM

Y by

We use the assertion $P(x,y)$ .

$P(0,0) \Longrightarrow f(0) = 0$ $P(0,y) \Longrightarrow f(-y)f(y) = f(y)^2$ Thus $f$ is even. $P(x,-x) \Longrightarrow 4f(y)^2 = 4 y^2 f(y)$ Thus either $f(y)\in \{0,y^2\}$ . Now we show that both cannot exist simultaneously. Let $\exists \; a,b \neq 0$ such that $f(a) = a^2$ and $f(b) = 0$ . $P(a,b) = P(b,a)$ since $f$ is even. Thus $P(a,b) = f(a+b)f(a-b) = f(a)^2 = P(b,a) = f(b+a)f(b-a) = f(a)^2 - 4b^2f(a)$ Thus $4b^2f(a) = 0$ but neither is $f(a) = 0$ nor is $b = 0$ . Contradiction. Thus the functions are $f(x) = 0$ or $f(x) = x^2$ .

This post has been edited 1 time. Last edited by SomeonecoolLovesMaths, Sep 28, 2024, 5:44 AM

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#39 h Sep 29, 2024, 3:38 PM

Y by

We use the proposition $P(x,y)$
$P(0,0):f(f(x))=f(0)$
$P(x,-f(x)): f(0)=f(f(x)+x^2)-4f(x)^2$
Now yes $x=0:$
$f(0)=f(f(0))-4f(0)^2$
$f(0)=f(0)-4f(0)^2 \Rightarrow f(0)=0$
Also $4f(x)^2=f(f(x)+x^2)$
$P(x,x^2): f(f(x)+x^2)= f(0)+4f(x)x^2=4f(x)x^2=4f(x)^2$
So $f(x)=0$ or $f(x)=x^2$
We can see that it is not possible for them to be mixed assuming that there are $a$ and $b$ other than zero such that $f(a)=0$ and $f(b)=b^2$ and arrive at $a$ and $b$ being defined as they already did in other solutions above.

This post has been edited 1 time. Last edited by MathIQ., Sep 29, 2024, 3:39 PM
Reason: typo

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#40 h Apr 25, 2025, 11:56 AM

Y by

$x=y=0\implies f(0)=0$ .
$x=0\implies f(-y)=f(y)$ .
$x=-y\implies 4y^2f(y)=4f(y)^2$ .
Now, say $f(a)=a^2$ and $f(b)=0$ , $b\neq0$ .
Since $f$ is even, $f(a+b)f(a-b)=f(a+b)f(b-a)$ .
Thus, $\left(f(a)+f(b)\right)^2=\left(f(a)+f(b)\right)^2-4b^2f(a)$ .
However, then $f(a)=0$ or $b=0$ , $\Rightarrow \Leftarrow$ .
Thus, $f\equiv 0$ or $f(x)=x^2$ .

This post has been edited 1 time. Last edited by Adywastaken, Yesterday at 3:25 PM

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