Find all real functions withf(x^2 + yf(z)) = xf(x) + zf(y)

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Find all real functions withf(x^2 + yf(z)) = xf(x) + zf(y)

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Source: INMO 2005 Problem 6

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1592 posts

Rushil

#1 h Aug 23, 2005, 1:21 PM • 3 Y

Y by mathematicsy, Adventure10, cubres

Find all functions $f : \mathbb{R} \longrightarrow \mathbb{R}$ such that $f(x^2 + yf(z)) = xf(x) + zf(y) ,$ for all $x, y, z \in \mathbb{R}$ .

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#2 h Aug 23, 2005, 1:49 PM • 2 Y

Y by Adventure10, Mango247

It is easy(ish) to see that $f$ is either identically zero or bijective. In either case we must have $f(0)=0$ .

In the latter case, suppose $f(z)=1$ , $x=0$ and we get $f(y)=zf(y)$ for all $y$ , so $f(1)=1$ .

Letting $x=0,y=1$ , we get $f(f(z))=z$ for all $z$ .

Letting $x=0,z=f(w)$ , we get $f(yf(f(w)))=f(yw)=f(y)f(w)$ for all $y,w$ , so $f$ is multiplicative.

Now for $x\ne0$ , $y=z=0$ , we get $f(x^2)=\left(f(x)\right)^2=xf(x)$ . Now since $f$ is bijective and $f(0)=0$ , $f(x)\ne0$ , and we get $f(x)=x$ .

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1592 posts

Rushil

#3 h Aug 23, 2005, 1:51 PM • 2 Y

Y by Adventure10, Mango247

I believe my solution is similar(probably there's only one) . Only a few different steps(effectively the same!!!

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#4 h Aug 23, 2005, 2:09 PM • 1 Y

Y by Adventure10

The INMO problems you've posted are quite nice.

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Saul

#5 h Sep 20, 2005, 8:56 AM • 2 Y

Y by Adventure10, Mango247

I have a different solution:

As Diarmuid said, it can easily be seen that if $f$ is not identically zero, it is surjective (by varying $z$ ). Bijectivity is non-essential, except at $x=0$ , which we will now prove. Suppose $f(z)=0$ for some $z \neq 0$ . Then we have $f(x^2)=xf(x)+zf(y)$ , a contradiction since $f(y)$ takes every real value.

Substituting $-x$ for $x$ gives that $xf(x)=-xf(-x)$ , and therefore that $f(x) = -f(-x)$ .

Now fix $x=y \neq 0$ and put $z = -x$ . We get that $f(x^2+xf(-x))=0$ and therefore $x^2=xf(x)$ (by bijectivity at 0) and $x=f(x)$ .

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Loser

#6 h Sep 21, 2005, 8:06 PM • 2 Y

Y by Adventure10, Mango247

Can you explain how do you get that $f$ is surjective, because I couldn't get it.
Also I didn't understand why $f$ is bijective.
Thanks in advance.

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Loser

#7 h Sep 26, 2005, 9:37 AM • 2 Y

Y by Adventure10, Mango247

Anyway, I understood the solution

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904 posts

#8 h Jan 15, 2010, 3:38 AM • 1 Y

Y by Adventure10

I have another (but strange

) solution to this problem...
Given is that $f(x^2 +yf(z)) = xf(x) +zf(y) \ \ \ \cdots (*)$
Taking $x=y=z=0$ ,
$f(0) = 0$
Now taking $y=z=0$
$f(x^2) = xf(x) \ \ \ \ \cdots \boxed{1}$

Putting $-x$ instead of $x$ in $\boxed{1}$ , we have
$f(x^2) = -xf(-x)$

Therefore, $xf(x) = -xf(-x) \iff f(-x) = -f(x) \ iff \ x \not = 0 \ \ \ \cdots \boxed{2}$
For the case $0$ , $\boxed{2}$ is also true, so it is true for all real $x$

Taking $y=z=k$ in $(*)$ , we have,
$f(x^2 +kf(k)) = xf(x) + kf(k)$
$\Rightarrow f(x^2 + f(k^2)) = f(x^2) + f(k^2) \ (From \ \boxed{2})$
Let $x^2 = a, k^2 = b$ with $a,b \ge 0$
$f(a+f(b)) = f(a) + f(b) \ \ \ \cdots \boxed{3}$

Now, taking $y=1, z \ge 0$ in $(*)$
$f(x^2 + f(z)) = xf(x) +zf(1)$
Therefore, by $\boxed{3}$ , we have
$f(x^2) + f(z) = xf(x) + zf(1)$
$\Rightarrow xf(x) +f(z) = xf(x) + zf(1)$ By $\boxed{1}$

Let $f(1) = c$ for some constant $c$
$\Rightarrow f(z) = cz \ \forall z \ge 0$
From $\boxed{2}$ ,
$f(-z) = -f(z) = c \cdot (-z)$ so $f(z) = cz$ is true for all real $z$ .

Substituting this information in $(*)$ , we get
$cx^2 + c^2yz = cx^2 +cyz$
$\Rightarrow c^2 = c$
$\Rightarrow c = 0,1$

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

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sayantanchakraborty

505 posts

sayantanchakraborty

#9 h Dec 22, 2013, 8:07 AM • 2 Y

Y by Adventure10, Mango247

$f(x^2+yf(z))=xf(x)+zf(y) (1)$

Taking y=z=0 in (1) we get
$f(x^2)=xf(x) (2)$
Taking x=0 in (2) we get f(0)=0
Taking x=0 and y=z=x in (1) we get
$f(xf(x))=xf(x) (3)$
Taking y=z=x in (1) we get,
$f(x^2+xf(x))=2xf(x) (4)$
Now using the fact that $f(x^2)=xf(x)$ we get
$f(x^2+xf(x))=f(x^2+f(x^2))=xf(x)+x^2f(1) (5)$
Comparing (4) and (5) we get,
$2xf(x)=xf(x)+cx^2$ where $c=f(1)$
or, $f(x)=cx$ considering x not equal to 0.
Substituting this back in (3) we get
$c^2x^2=cx^2$
or, $c^2=c$
From this we get that c=0 or 1 and consequently
$f(x)=0$ or
$f(x)=x$ for all real x

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#10 h Jan 27, 2014, 10:32 AM • 1 Y

Y by Adventure10

Somebody, please check my solution:
$f(x^2+yf(z))=xf(x)+zf(y)$
Let $P(x,y,z)$ be the assertion for the functional equation.
$P(0,0,0)\implies f(0)=0$
$P(x,0,0)\implies f(x^2)=xf(x)$
$\implies f(x^2+yf(z))=x^2+zf(y) \cdots \star$
$P(0,y,z)\implies f(yf(z))=zf(y)$
$\implies f(x^2+yf(z))=f(x^2)+f(yf(z))$
Which is the Cauchy's functional equation and has solutions $f(x)=cx,f(x)=0$
from here we can prove $c=1$ as Sayantan(previous poster) did and this gives us solutions $f(x)=x \text{ or }f(x)=0\forall x\in \mathbb{R}$
$%Error. "blackbox" is a bad command.$
From the equation marked with a star we may also proceed as follows:
We can prove $f(x)$ is injective except the solution $f(x)=0\forall x\in \mathbb{R}$
Now $P(x,y,-\frac{x^2}{f(y)})\implies f(x^2+yf(-\frac{x^2}{f(y)}))=0$
As the function is injective for the non-zero solutions, also $f(0)=0$ ,
We have,
$x^2+yf(-\frac{x^2}{f(y)})=0\implies f(-\frac{x^2}{f(y)})=-\frac{x^2}{y}$
Hence the solution $f(x)=x\forall x\in \mathbb{R} \text{ or } f(x)=0\forall x\in \mathbb{R}$

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331 posts

#11 h Jan 28, 2014, 12:00 AM • 1 Y

Y by Adventure10

Yet another solution:

Let $P$ be the given assertion. Then we get:
$P(0, 0, 0)\implies f(0)=0$
$P(x, 0, 0)\implies f(x^2)=xf(x)$
$P(0, x, x)\implies f(xf(x))=xf(x)$
$P(0, 1, x)\implies f(f(x))=xf(1)$

Then from the statements above, we have:
$x^2f(1)=f(f(x^2))=f(xf(x))=xf(x)\implies f(x)=xf(1)$ .
$P(1, 1, 1)\implies f(1)+f(1)^2=f(1+f(1))=2f(1)\implies f(1)^2=f(1)$

Therefore, $f(1)=0$ or $1$ , so:
$f(x)=0$ or $f(x)=x$ .

Seriously? This was InMO #6?

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#12 h Jul 28, 2014, 3:52 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Ashutoshmaths wrote:

$P(x,0,0)\implies f(x^2)=xf(x)$
$\implies f(x^2+yf(z))=x^2+zf(y) \cdots \star$

I think I found a mistake
It probably should imply $f(x^2+yf(z))=f(x^2)+zf(y)$

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#13 h Sep 14, 2014, 2:50 PM • 1 Y

Y by Adventure10

I have got a solution apparently different

Please confirm if it is correct !

$f(x^2+yf(z))=xf(x)+zf(y)$
Let $P(x,y,z)$ be the assertion of the functional equation

$P(x,0,0) \implies f(x^2)=xf(x)$

$f(x)=0$ clearly satisfies the equation.
Now let $f(x)$ be an identically non zero function.

$P(0.y.z) \implies f(yf(z))=zf(y)$

Then $f(f(yf(z)))=f(zf(y))=yf(z)$
For each $x$ we can chose $y,z$ appropriately (since it is a non zero function) such that,
$yf(z)=x$

Thus $f(f(x))=x$ that is $f(x)$ is involutive

Also $xf(x)=f(xf(x))$ (Since $f(yf(z))=zf(y)$ as above)

Combining these results
$xf(x)=f(xf(x))=f(f(x^2))=x^2$

That is $f(x)=x$ for all $x$

Thus the only solutions are
$f(x)=0$ and $f(x)=x$

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804 posts

Sardor

#14 h Sep 14, 2014, 3:39 PM • 2 Y

Y by Adventure10, Mango247

It's easy and nice function.Here my solution:
Let $P(x,y,z)$ be the assertion of the function.
1) $P(0,0,0) \implies f(0)=0$ .
2) $P(x,0,0) \implies f(x^2)=xf(x) (*)$ and $P(0,1,z) \implies f(f(z))=f(1)z$ .It's easy to see that $f$ is zero function or bijective fuction.In least identity we take $z=1 \implies f(f(1))=f(1)$ so by bijectivite we get $f(1)=1$ and so $f(f(z))=z (**)$ .
3) From $(*)+(**)$ we have :
$f(f(x)^2)=f(x)f(f(x))=f(x)x=f(x^2) \implies f(x)=x,-x$ . But we see that $f(x)=-x$ isn't solution.
Answers:
1) $f(x)=0$
2) $f(x)=x$

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1566 posts

#15 h Dec 27, 2017, 8:02 AM • 2 Y

Y by Adventure10, Mango247

We claim that $f(x)=0$ $\forall x \in R$ or $f(x)=x$ $\forall x \in R$ are the only two solutions. It is easy to check that these 2 work. We now show that these are the only 2 solutions.
$[1]$ Set $x=y=z=0$ to get $f(0)=0$ .
$[2]$ Set $y=0$ to get $f(x^{2})=xf(x)$ $\forall x \in R$ . Thus the equation now takes the form $f(x^{2}+yf(z))=f(x^{2})+zf(y)$ .
$[3]$ Set $x=0$ to get $f(yf(z))=zf(y)$ $\forall y,z \in R$ .
$[4]$ Set $x=-x$ in $[2]$ to get $f(-x)=-f(x)$ .
$[5]$ Set $x=\sqrt{y(1-f(z))}$ in $[2]$ to get (according to the sign of $1-f(z)$ , we choose the sign of $y$ to ensure that $x \in R$ , and hence either $y \in R^{+}$ , or $y \in R^{-}$ . We do not consider $y=0$ .)
$f(y)=f \left( \left (\sqrt{y(1-f(z))} \right)^{2} +yf(z) \right)=f(y(1-f(z)))+zf(y) \Leftrightarrow (1-z)f(y)=f \left( y(1-f(z)) \right)$ In this equation, set $z=1$ to get $f \left( y(1-f(1)) \right)=0$ $\forall y \in R^{+}$ or $R^{-}$ , depending on whether $1 \geq f(1)$ or $1 < f(1)$ . Since this is true for all $R^{+}$ (or $R^{-}$ ), either $1-f(1)=0$ or $f(x)$ is a constant on one of these 2 domains. In the second case, we can use $[4]$ and hence easily get $f(x)=0$ $\forall x \in R$ .
So now assume $f(1)=1$ . This is our result $[6]$ .
$[7]$ Now, substituting $y=1$ in $[3]$ and using $[6]$ we get $f(f(y))=y$ $\forall y \in R$ .
Hence $f(f(x^{2}))=x^{2}$ $\forall x \in R$ . But setting $y=z$ in $[3]$ (and using $[2]$ ) gives $f(f(y^{2}))=f(y^{2})$ $\forall y \in R$ . Thus, $f(a)=a$ $\forall a \in R^{+}$ . Then using $[4]$ gives $f(x)=x$ $\forall x \in R$ . (We can check that this is true for the case $x=0$ as well).

This post has been edited 1 time. Last edited by Wizard_32, Dec 27, 2017, 8:03 AM
Reason: errrrrrrrr

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#16 h Mar 11, 2020, 1:06 PM • 1 Y

Y by amar_04

Amusing one for showing for the positive reals.

Trivial to note $f(0)=0$ , then $P(x,y,0)$ gives $f(x^2)=xf(x)$ so that $P(f(x),y,y)$ gibes $f(f(x)^2+f(y)^2)=f(x)^2 + f(y)^2$ and varying $x,y$ we get the result for positive reals

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#17 h Oct 19, 2020, 2:31 PM • 1 Y

Y by Mango247

Allzero works, so assume henceforth that $\exists u$ , $f(u)\neq 0$
$P(1,u,z)\implies f(1+uf(z))=f(1)+zf(u)$
$P(0,0,0)\implies f(0)=0 \implies u\neq 0 \implies$ bijective
$P(x,0,0)\implies f(x^2)=xf(x)$ and $P(0,x,x)\implies f(xf(x))=xf(x)$
Hence, $f(x^2)=f(xf(x))$ and since bijective, $x^2=xf(x) \implies f(x)=x, \forall x\neq0$
So, $f(x)=x, \forall x\in\mathbb{R}$ .
Hence two solutions:
$f(x)=0, \forall x\in\mathbb{R}$ and $f(x)=x \forall x\in\mathbb{R}$

This post has been edited 3 times. Last edited by lahmacun, Oct 19, 2020, 2:32 PM

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#18 h Oct 21, 2020, 1:16 PM

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clearly, if $f$ is constant then it's zero so suppose that it's not constant
let $P(x,y)$ be the assertion of the equation.
$P(0,y,z)\implies f(yf(z))=zf(y)$ so $f(xf(x))=xf(x)$
subbose that $f(a)=f(b)$ for some $a,b$ and compare $P(x,y,a)$ with $P(x,y,b)$
$\implies f$ is bejective
now , $P(x,0,0)\implies f(x^2)=xf(x)=f(xf(x))$
so $\boxed{f(x)=x \quad \text{or}\quad 0 \quad \forall x\in\mathbb{R}}$

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#19 h Apr 1, 2021, 3:07 PM

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Answer: $f(x) = x \quad \text{or} \quad f(x) = 0$

Proof:
Let $P(x,y,z)$ denote the given assertion.
$P(0,0,0) \to f(0) = 0$
$P(x,0,0) \to \color{red}{f(x^2) = xf(x)}$
$P(x,x,x) \to f(x^2 + xf(x)) = xf(x) + xf(x)$
$\implies \color{red}{f(x^2 + f(x^2)) = 2xf(x)}$
$P(x,1,x^2) \to f(x^2 + f(x^2)) = xf(x) + x^2f(1)$
$\implies 2xf(x) = xf(x) + x^2f(1)$
$\implies xf(x) = x^2f(1)$
$\implies f(x) = xf(1)$ (Holds even when $x=0$ )
If $f(1)=c \implies f(x) = cx$
Plugging this back into our original equation.
$c(x^2 + cyz) = cx^2 + cyz$
$\implies c^2 = c$
$\implies c=0,1$
After checking we see that both values of $c$ work.
Hence $f(x) = x \quad \text{or} \quad f(x) = 0$ .

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#20 h Apr 1, 2021, 3:14 PM

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$P(x,0,0)\Rightarrow f(x^2)=xf(x)\Rightarrow f$ is odd.
$P(x,y,1)\Rightarrow f(x^2+y)=f(x^2)+f(y)$ so since $f$ is odd, it is additive.
This is now USAMO 2002 P4.

Alternatively:
$P(0,x,y)\Rightarrow Q(x,y):f(xf(y))=yf(x)$
$Q(1,x)\Rightarrow f(f(x))=xf(1)$
If $f(1)=0$ then $Q(x,f(x))\Rightarrow f(x)^2=0$ , so $\boxed{f(x)=0}$ , which works.
Otherwise, $f$ is bijective.
$Q(1,1)\Rightarrow f(f(1))=f(1)\Rightarrow f(1)=1$
$Q(1,x)\Rightarrow f(f(x))=x$
$Q(x,f(y))\Rightarrow f(xy)=f(x)f(y)$ and since $f$ is additive and multiplicative, $f(x)=cx,c\in\{-1,1\}$ . Checking, we see that $\boxed{f(x)=x}$ .

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jelena_ivanchic

151 posts

jelena_ivanchic

#21 h Jan 5, 2022, 5:50 AM • 3 Y

Y by Pranav1056, Jupiter_is_BIG, SatisfiedMagma

Oh, I really overcomplicated it :O But yeyy, I solved an INMO fe!

Let $P(x,y)$ denote the assertion.
Then $P(x,0,0)\implies f(x^2)=xf(x)\implies f(x)=-f(-x)$
And $f(x^2+y^2)=f(x^2)+f(y^2)\implies f(a+b)=f(a)+f(b), a,b \in \mathbb{R}^{+}$
Also $f(0)=0.$ Now we show that $f$ is injective at $0.$ So let $f(t)=0, t\ne 1.$ So $P(0,1,t)\implies 0=f(f(t))=tf(1)\implies f(1)=0\text{ or } f=0$ Assuming $f$ is non constant, we get $f(1)=0.$ Then $P(0,a,1)\implies 0=f(a)\forall a.$
Not possible, hence $f$ is injective at $0.$
Moreover, $P(0,1,x)\implies f(f(x))=xf(1)$
Note that $f(f(x))=xf(1)\implies f \text { is bijective }$
Actually, we don't need the following claim, but I found it too good to not add.

Claim: $f(1)=1$
Proof: $P(0,y,1)\implies f(yf(1))=f(y)\implies yf(1)=y\implies f(1)=1$
Alternate proof: Using subjectivity, $\exists k, f(k)=1\implies P(0,k,k)\implies 1=f(kf(k))=k\implies k=1$
Now, fix $x.$ We carry $y$ and choose $z$ such that $f(x^2+yf(z)=xf(x)+zf(y)=0$ Then $x^2+yf(z)=0, xf(x)+zf(y)=0$
Take $y=x\implies z=-x$ in $xf(x)+zf(y)=0.$ So, we get $x^2+xf(-x)=0\implies f(-x)=-x\implies f(x)=x$
Hence the solutions are $\boxed{ f=0,f(x)=x}$

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ZETA_in_olympiad

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ZETA_in_olympiad

#22 h Jun 29, 2022, 11:23 AM

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Denote the assertion by $P(x,y,z).$ Clearly $f(0)=0.$ Then $P(0,1,x)$ gives $f(f(x))=cx$ so $f$ is bijective otherwise if $c=0$ then $P(0,x,f(x))$ gives $f(x)\equiv 0$ this works. Now comparing $P(x,x,x)$ and $P(x,0,x)$ we get $f(x)\equiv x,$ which works.

This post has been edited 3 times. Last edited by ZETA_in_olympiad, Jun 29, 2022, 11:35 AM

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#23 h Jun 29, 2022, 11:57 AM

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

$P(0,x,y): f(xf(y))=yf(x)$ .

$P(0,1,x): f(f(x))=xf(1)$ .

If $f(1)=0$ , then $f(f(x))=0$ , $P(0,x,f(y))$ gives $f(0)=f(x)f(y)$ . Setting $y=1$ gives $f(0)=f(x)f(1)=0$ . So $f(x)f(y)=0\implies f(x)^2=0\implies \boxed{f\equiv 0}$ .

Henceforth assume $f(1)\ne 0$ . Then $f$ is bijective.

Note that $f(f(1))=f(1)$ , so $f(f(f(1)))=f(f(1))=f(1)$ .

$P(0,1,f(1)): f(1) = f(1)^2\implies f(1)=1$ since $f(1)\ne 0$ .

Thus, $f(f(x))=x$ .

$P(0,x,f(y)): f(xy)=f(x)f(y)$ , so $f$ is multiplicative.

$P(x,y,0): f(x^2)=xf(x)$ , so $f$ is odd.

$P(x,y,1): f(x^2+y)=f(x^2)+f(y)$ . Since $f$ is odd, $f$ is additive.

Now, $f$ is additive and multiplicative, so $f(x)=cx$ for some constant $c\ne 0$ , checking gives $\boxed{f(x)=x}$ or $\boxed{f(x)=-x}$ , which work.

This post has been edited 1 time. Last edited by megarnie, Jun 29, 2022, 11:57 AM

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F10tothepowerof34

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F10tothepowerof34

#25 h Apr 20, 2023, 10:34 AM

Y by

Claim:
$\boxed{f(x)=cx, \forall x\in \mathbb{R}}$ where $c=0$ or $c=1$

Proof:
Let $P(x,y,z)$ denote the assertion: $f(x^2+yf(z))=xf(x)+zf(y)$
$P(0,0,0)$ yields $f(0)=0$
$P(x,0,0)$ yields $f(x^2)=xf(x)$
Furthermore $P(x,x,x)\implies f(x^2+f(x^2))=2xf(x)$ (1)
$P(x,1,x^2)$ yields $f(x^2+f(x^2))=xf(x)+x^2$
However notice that from (1) we have:
$2xf(x)=xf(x)+x^2f(1)\Longrightarrow f(x)=xf(1)$
Thus, since $f(1)$ is a constant, we have that $f(x)=cx$
$P(1,1,1)$ yields $f(c+1)=2c\Longrightarrow c^2=c$
Thus $c$ can only be equal to $1$ or $0$ .QED
And we are done!

This post has been edited 3 times. Last edited by F10tothepowerof34, Apr 23, 2023, 8:02 PM

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egxa

#26 h Apr 20, 2023, 11:41 AM • 2 Y

Y by AlperenINAN, bin_sherlo

Let $P(x,y,z)$ be the assertion $f(x^2+yf(z))=xf(x)+zf(y)$
$P(0,0,0)$ gives $f(0)=0$
$P(x,y,0)$ gives $f(x^2)=xf(x)$
$P(x,y,c_1)$ and $P(x,y,c_2)$ gives $f$ is injective or $f(x)=0$
$P(0,x,x)$ gives $f(xf(x))=xf(x)$ so $f(f(x^2))=f(xf(x))=xf(x)=f(x^2) f(x^2)=x^2$ and from $f : \mathbb{R} \longrightarrow \mathbb{R}$ solutions are $f(x)=x$ and $f(x)=0$

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#27 h Apr 28, 2023, 9:20 AM

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Very similar to IMO 1992/2. We use pretty much the same idea here too!

Solution. The answers are $f \equiv 0$ and $f \equiv \mathrm{id}$ . All of them work. Denote the assertion to the functional equation by $P(x,y,z)$ . We will now proceed towards to the other direction.

Its clear that $f \equiv 0$ is the only constant. $P(0,0,0)$ reveals $f(0) = 0$ . If $f$ attains $0$ at any non-zero constant $\alpha$ , then $P(0,y,\alpha)$ would yield $f \equiv 0$ . Henceforth, assume that $f$ is non-zero everywhere except $0$ .

$P(x,0,0)$ gives $f(x^2) = xf(x)$ . $P(0,y,1)$ gives us $f(yf(1)) = f(y) \implies f(1) = 1$ and finally $P(0,1,z)$ gives us that $f$ is bijective and an involution. Here's the nice part, $P(x,y,1)$ gives
$f(x^2 + y) = xf(x) + f(y) = f(x^2) + f(y) \implies f(a+b) = f(a) + f(b)$ for all real $a$ and $b \ge 0$ . This condition is well-known over $\mathbb{R}$ and can be extended to full a Cauchy. $P(0,y,f(y))$ would give $f(y^2) = f(y)^2$ . So, $f > 0$ for positive values of domain which due to Cauchy means that $f$ is strictly increasing. Since $f$ is strictly increasing and $f$ is additive, $f(x) = ax$ for some real constant $a$ . Just checking it once, we're done. $\blacksquare$

This post has been edited 2 times. Last edited by SatisfiedMagma, Apr 28, 2023, 9:21 AM
Reason: error

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#28 h Dec 9, 2023, 9:00 AM

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Claim 1: $f(0)=0$
Proof: trivial
Claim 2: $f(x^2)=xf(x)$
Proof: Putting $y=z=0 \implies f(x^2)=xf(x)$
Claim 3: $f$ is an involution
Proof: Putting $x=0 \implies f(yf(z))=zf(y) \implies f(f(yf(z))=f(zf(y))=yf(z)$
But $yf(z)$ is surjective in $\mathbb{R}$
Therefore $f(f(x))=x \forall x \in \mathbb{R}$
Claim 3: $f$ is odd
Proof: Putting $x=f(z), y=-f(z)$ in original equation
We get $f(f(z)^2-f(z)^2)=f(0)=0=f(z)f(f(z))+zf(-f(z))=f(z)z+zf(-f(z)) \implies f(-f(z))=-f(z)$ now since $f$ is surjective therefore $f$ is also odd.
Claim 4: f(x)=x
Proof: taking $y=z$
We get $f(x^2+y^2)=x^2+y^2 \implies f(x)=x$ for all positive real x. But since $f$ is odd therfore we can extend this to the negative numbers as well and hence we are done.

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SouradipClash_03

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#29 h Jan 5, 2024, 5:16 PM

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How was this INMO P $6$ wth. Just tried out random stuffs and got my solution.
$P(0,0,0) \implies f(0) = 0$ .
$P(x,0,0) \implies f(x^2) = xf(x)$ .
Comparing $P(x,0,0)$ with $P(-x,0,0)$ gives us $f(x) = -f(-x)$ , so $f$ is odd.
Now, consider $a,b \in \mathbb{R}$ where $f(a) = f(b)$ .
Then, $xf(x) + af(y) = xf(x) + bf(y)$
Clearly $f(x) = 0$ is a solution, then assume $f(x) \neq 0$ .
This gives out $a = b$ , and thus $f$ is injective.
$P(0,y,1) \implies f(yf(1)) = f(y)$ .
$\implies yf(1) = y$ .
$\implies f(1) = 1$ .
Now, $P(0,1,z) \implies f(f(z)) = z$ .
So, $f$ is involutive.
$P(0,x,x) \implies f(xf(x)) = f(x^2) = xf(x)$ [From $xf(x) = f(x^2)$ ]
Or, $f(x) = x$ .
Overall, we get $2$ solutions, $f(x) = 0$ and $f(x) = x$ , both seem to work.

This post has been edited 4 times. Last edited by SouradipClash_03, Jan 5, 2024, 5:40 PM

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#30 h Jan 5, 2024, 5:59 PM • 1 Y

Y by ATGY

Really interesting problem.

We claim the only solutions are $f(x) \equiv 0 \forall x$ and $f(x) = x \forall x$ . Clearly, they satisfy the given equation. We now prove these are the only solutions.
Since $f(x) \equiv 0 \forall x$ satisfies, the equation, assume $f$ is not identically zero, and let $P(x, y, z)$ denote the assertion $f(x^2 + yf(z)) = xf(x) + zf(y)$ .
$P(0, 0, 0): f(0) = 0$
$P(0, y, z): f(yf(z)) = zf(y) \implies f(f(yf(z))) = yf(z)$ .
Note that we can choose z such that $f(z) \ne 0$ , and therefore $yf(z)$ can represent any real, say $a$ .
Therefore we have $f(f(a)) = a$ , i.e. $f$ is an involution and therefore bijective.
Now $P(0, y, 1): f(yf(1)) = f(y) \implies yf(1) = y \implies f(1) = 1$ .
Now, $P(x, y, 1): f(x^2 + y) = xf(x) + f(y)$ (Denote this by (1))
Now note that $f$ is odd, as $P(x, 0, 0): f(x^2) = xf(x) = (-x)f(-x)$ .
Therefore (1) simplifies as $f(x^2 + y) = f(x^2) + f(y)$ .
Now since this holds for all real $x$ and $y$ , it holds for all positive reals $x, y$ , so we assume $x, y>0$ , and let $b = x^2$ .
Then we have $f(b+y) = f(b) + f(y)$ , and since we are assuming f to be bounded (in the positive reals) , this satisfies Cauchy's Functional Equation, i.e. we get $f(x) = cx \forall$ positive real x.
Also, $f(-x) = -f(x) = -cx = c(-x)$ for positive $x$ , so $f(x) = cx$ for all nonzero x, and as $f(0) = 0$ , we get $f(x) = cx \forall x$ .
Substituting into the original equation, we get $cx^2 + c^2 yz = cx^2 + cyz \implies c^2 = c,$ and as $c \ne 0$ we have $c = 1$ , i.e. $f(x) = x$ .
Therefore, $f(x) \equiv 0$ and $f(x) = x$ are the only solutions.

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#31 h Jan 10, 2024, 3:10 PM

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Bruh

First notice that $f(x)=0$ obviously works, now assume that $f(x)$ is nonzero for atleast one input value, say $c \neq 0$
$P(0,0,0) \implies f(0) = 0$ .
Now $P(0,c,z) \implies f(cf(z))=zf(c)$ which implies that $f$ is bijective. Finally $P(x,1-x,x) \implies f(x^2+xf(1-x))=f(x) \implies x^2+xf(1-x)=x \implies f(1-x)=1-x \implies f(x)=x \ \ \forall \ x \in R-0$ and since we know that $f(0)=0$ we can conclude that $f(x)=0$ and $f(x)=x$ are the only solutions $\blacksquare$

Also, INMO doesn't order its problems difficulty-wise, so this being p6 (in 2005) is not a big deal

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SomeonesPenguin

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#34 h Aug 13, 2024, 2:14 PM • 1 Y

Y by zzSpartan

A little boring.

Solution

This post has been edited 1 time. Last edited by SomeonesPenguin, Aug 13, 2024, 2:16 PM

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#35 h Apr 15, 2025, 7:03 PM

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The answers are $f(x) \equiv 0$ and $f(x) = x$ . Let $f(x;y;z) = f(x² + yf(z)) = xf(x) + zf(y)$ . So $f(0;0;0) \implies f(0) = 0$ , $f(0;y;z) \implies f(x²) = xf(x)$ and $f(0;y;z) \implies f(yf(z)) = zf(y)$ , so $f(0;x;x) \implies f(xf(x)) = xf(x) = f(f(x²)) = f(x²)$ , $f(0;1;x) \implies f(f(x)) = xf(1)$ , so $f(f(xf(x))) = f(xf(x)) = xf(x) = xf(x)f(1) \implies xf(x)(1 - f(1)) = 0 \implies$ for $x \neq 0, f(1) = 1$ or $f(x) = 0 \ \ \forall x$ . If $f(x) \equiv 0$ , we're done with this case. If $f(1) = 1 \implies f(f(x)) = x \implies f(f(x²)) = f(x²) = x²$ , so for all $x \geq 0$ , $f(x) = x$ . Let x be a negative real number, so $f(x²) = xf(x) = x² \implies f(x) = x$ for x negative, so we conclude that $f(x) = x$ is the other case and we're done.

This post has been edited 3 times. Last edited by Jakjjdm, Apr 15, 2025, 7:08 PM

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#36 h Apr 15, 2025, 9:36 PM

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solution

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