IMO Shortlist 2009 - Problem A7

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April

1270 posts

April

#1 h Jul 5, 2010, 11:32 AM • 7 Y

Y by Mathuzb, itslumi, megarnie, Stuffybear, Adventure10, Mango247, and 1 other user

Find all functions $f$ from the set of real numbers into the set of real numbers which satisfy for all $x$ , $y$ the identity $f\left(xf(x+y)\right) = f\left(yf(x)\right) +x^2$

Proposed by Japan

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Bacteria

256 posts

Bacteria

#2 h Jul 8, 2010, 12:58 AM • 1 Y

Y by Adventure10

Here is my quite lengthy solution:
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daniel73

253 posts

daniel73

#3 h Jul 9, 2010, 7:29 AM • 3 Y

Y by Adventure10, Mango247, Gaunter_O_Dim_of_math

I try to shorten it a little bit:

1st part, we show $f(0)=0$ , $f$ is bijective and $f(x)=-f(-x)$ :
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2nd part, we prove that for all $z,x$ , we have Click to reveal hidden text
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3rd part, conclusion
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tenniskidperson3

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tenniskidperson3

#4 h Jul 17, 2010, 5:46 PM • 2 Y

Y by Adventure10, Mango247

I don't understand how $x^{2}=f(xf(z))-f((z-x)f(x))$ and $f(x)=f(z)$ implies $z-x=0$ or $x=0$ . Could you explain?

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daniel73

253 posts

daniel73

#5 h Jul 19, 2010, 11:56 PM • 2 Y

Y by Adventure10, Mango247

tenniskidperson3 wrote:

I don't understand how $x^{2}=f(xf(z))-f((z-x)f(x))$ and $f(x)=f(z)$ implies $z-x=0$ or $x=0$ . Could you explain?

Since $f(z)=f(x)$ , then $f(xf(z))=f(xf(x))=x^2$ , or $f((z-x)f(x))=0$ , ie by previous results $(z-x)f(x)=0$ , yielding either $z=x$ or $f(x)=0$ , this latter one being equivalent again to $x=0$ .

Hope this clears it out...

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lefao

150 posts

lefao

#6 h Aug 2, 2010, 5:08 AM • 2 Y

Y by Adventure10, Mango247

daniel73 wrote:

tenniskidperson3 wrote:

I don't understand how $x^{2}=f(xf(z))-f((z-x)f(x))$ and $f(x)=f(z)$ implies $z-x=0$ or $x=0$ . Could you explain?

Since $f(z)=f(x)$ , then $f(xf(z))=f(xf(x))=x^2$ , or $f((z-x)f(x))=0$ , ie by previous results $(z-x)f(x)=0$ , yielding either $z=x$ or $f(x)=0$ , this latter one being equivalent again to $x=0$ .

Hope this clears it out...

What if $f((z-x)f(x)=x^2$ ? It also imply $z=x$ or $x=0$ ?

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daniel73

253 posts

daniel73

#7 h Aug 2, 2010, 6:08 AM • 2 Y

Y by Adventure10, Mango247

lefao wrote:

daniel73 wrote:

tenniskidperson3 wrote:

I don't understand how $x^{2}=f(xf(z))-f((z-x)f(x))$ and $f(x)=f(z)$ implies $z-x=0$ or $x=0$ . Could you explain?

Since $f(z)=f(x)$ , then $f(xf(z))=f(xf(x))=x^2$ , or $f((z-x)f(x))=0$ , ie by previous results $(z-x)f(x)=0$ , yielding either $z=x$ or $f(x)=0$ , this latter one being equivalent again to $x=0$ .

Hope this clears it out...

What if $f((z-x)f(x)=x^2$ ? It also imply $z=x$ or $x=0$ ?

You have a bracket missing; if you mean $f((z-x)f(x))=x^2$ , then we know by previous results that $x^2=f(xf(x))$ , and that $f$ is injective, or you may deduce that $f(z-x)f(x)=xf(x)$ , yielding either $f(x)=0$ (which again results in $x=0$ ), or $f(z-x)=x$ . Since we have chosen $x\neq0$ , you would obtain that $f((z-x)f(x))=x^2$ implies $f(z-x)=x$ . However, I do not know under which circumstances you would obtain $f((z-x)f(x))=x^2$ , and I do not see clearly why $f(z-x)=x$ would be useful, but if it is, feel free to use it.

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Zhero

2043 posts

Zhero

#8 h Aug 12, 2010, 4:24 AM • 3 Y

Y by Tawan, alinazarboland, Adventure10

Set $x = 0$ . $f(0) = f(yf(0))$ for all $y$ . If $f(0) \neq 0$ , $yf(0)$ spans the set of reals, so $f$ must be constant, which is clearly impossible. Hence, $f(0) = 0$ .

Setting $y=0$ shows that $f(xf(x)) = x^2$ for all real $x$ , and setting $x=-y$ shows that $f(-xf(x)) = -x^2$ for all real $x$ . It follows from these that if $f(v) = 0$ , then $v^2 = f(vf(v)) = f(0) = 0$ , i.e., $v = 0$ .

We now claim that $f$ is injective. Suppose that $f(a) = f(a+b)$ . We wish to show that $b=0$ . From $f(af(a)) = a^2$ , we have $a^2 = f(af(a)) = f(af(a+b)) = f(bf(a)) = a^2$ , whence $f(bf(a)) = 0$ . Hence, $b = 0$ or $f(a) = 0$ . If $f(a) = 0$ , then from the above $a = b = 0$ , so $f$ is injective.

From $f(xf(x)) = x^2$ , we have $xf(x) f(xf(x)) = x^3 f(x)$ , so $f(x^3 f(x)) = f((xf(x)) f(xf(x))) = x^2 f(x)^2$ . When $x=1$ , we have $1^2 = f(1 f(1)) = f(1^3 f(1)) = f(1)^2$ . Similarly, $f(-1)^2 = 1$ . By injectivity, either $f(1) = 1$ and $f(-1) = -1$ , or $f(-1) = 1$ and $f(1) = -1$ . Since $f(x)$ is a solution to the functional equation if and only if $f(-x)$ is (as we may replace each occurence of $x$ and $y$ in the original functional equation with $-x$ and $-y$ ), we may for now assume without loss of generality that $f(1) = 1$ and $f(-1) = -1$ .

Setting $y=x$ in the original functional equation gives $f(xf(2x)) = f(xf(x)) + x^2 = 2x^2$ . Since $f(\sqrt{2} x f( \sqrt{2} x)) = 2x^2$ as well, by injectivity $x f(2x) = \sqrt{2} x f( \sqrt{2} x)$ . Replacing $x$ with $\frac{\sqrt{2} x}{2}$ and rearranging gives $f(\sqrt{2} x) = \sqrt{2} f(x)$ . Hence, $f(2x) = f( \sqrt{2} (\sqrt{2} x)) = \sqrt{2} f(\sqrt{2} x) = 2x$ .

Setting $x=1$ in the original functional equation gives $f(f(y+1)) - f(y) = 1$ , and setting $x=2$ gives $f(f(y+2)) - f(y) = 2$ . Subtracting these two equations gives $f(f(y+2)) - f(f(y+1)) = 1$ for all $y$ , or $f(f(y+1)) = f(f(y)) + 1$ . Substituting this into $f(f(y+1)) - f(y) = 1$ yields $f(f(y)) = f(y)$ . By injectivity, $f(y) = y$ for all real $y$ .

Recall that this is only the solution when $f(1) = 1$ ; if $f(1) = -1$ , $f(-x)$ is the solution to this functional equation, so $f(x) = x$ and $f(x) = -x$ are the only solutions.

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orl

3647 posts

orl

#9 h Oct 16, 2010, 11:53 PM • 6 Y

Y by 62861, Tawan, HouseofTerror_43, megarnie, Adventure10, khina

Information by Kazuaki Kobayashi, MOF (Mathematic Olympiad Foudation) of Japan about the background for this problem.

Attachments:

SL2009 A7.pdf (15kb)

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pco

23515 posts

pco

#10 h Dec 21, 2010, 2:13 PM • 5 Y

Y by Tawan, MeowX2, Adventure10, Mango247, and 1 other user

bigbang195 wrote:

Find all functions $f$ from the set of real numbers into the set of real numbers which satisfy for all $x$ , $y$ the identity $f\left(xf(x+y)\right) = f\left(yf(x)\right) +x^2$

Proposed by Japan

Here is a long and rather ugly solution.

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

1) $f(x)=0$ $\iff$ $x=0$
==============
If $f(0)\ne 0$ , then $P(0,\frac x{f(0)})$ $\implies$ $f(x)=f(0)$ constant, which is not a solution. So $f(0)=0$
If $f(u)=0$ , then $P(u,-u)$ $\implies$ $u^2=0$
Q.E.D.

2) $f(x)$ is a bijection
=============
$P(x,0)$ $\implies$ $f(xf(x))=x^2$ and so $\mathbb R^+\cup\{0\}\subseteq f(\mathbb R)$
$P(x,-x)$ $\implies$ $f(-xf(x))=-x^2$ and so $\mathbb R^-\cup\{0\}\subseteq f(\mathbb R)$
And so $f(x)$ is a surjection

Let then $a<0$ and $u$ such that $f(u)=a$ :
$P(\sqrt{-a},\frac u{f(\sqrt{-a})})$ $\implies$ $f(\sqrt{-a}f(\sqrt{-a}+\frac u{f(\sqrt{-a})}))=0$
So (using 1. above) : $f(\sqrt{-a}+\frac u{f(\sqrt{-a})})=0$
So (using again 1) above) : $\sqrt{-a}+\frac u{f(\sqrt{-a})}=0$ and $u=-\sqrt{-a}f(\sqrt{-a})$
And so a unique $u$ for any $a<0$

Let then $a>0$ and $u\ne 0$ such that $f(u)=a$ and $v$ such that $f(v)=\frac u{\sqrt a}$ :
$P(\sqrt a,v-\sqrt a)$ $\implies$ $f((v-\sqrt a)f(\sqrt a))=0$
So (using 1. above) : $(v-\sqrt a)f(\sqrt a)=0$ and so $v=\sqrt a$ and $u=\sqrt af(\sqrt a)$
And so a unique $u$ for any $a>0$

And so $f(x)$ is an injection.
Q.E.D.

3) $f(x)$ is an odd function
=================
Let $x\ne 0$
$P(x,0)$ $\implies$ $f(xf(x))=x^2$
$P(-x,0)$ $\implies$ $f(-xf(-x))=x^2$
And so $f(xf(x))=f(-xf(-x))$ and so, since injective, $xf(x)=-xf(-x)$ and $f(-x)=-f(x)$ (true also if $x=0$ )
Q.E.D.

4) $f(2x)=2f(x)$
=========
Let $x\ne 0$ :
$P(x,-2x)$ $\implies$ $-f(xf(x))=-f(2xf(x))+x^2$
$P(-x,-x)$ $\implies$ $f(xf(2x))=f(xf(x))+x^2$
Subtracting, we get : $f(xf(2x))=f(2xf(x))$ and, since injective : $xf(2x)=2xf(x)$ and so $f(2x)=2f(x)$ (true also if $x=0$ )
Q.E.D.

5) $f(x+y)=f(x)+f(y)$ $\forall x,y$
=====================
$P(x,y)$ $\implies$ $f(xf(x+y))=f(yf(x))+x^2$
$P(x+y,-x)$ $\implies$ $f((x+y)f(y))=-f(xf(x+y))+(x+y)^2$
Subtracting, we get a new assertion $Q(x,y)$ : $f((x+y)f(y))+f(yf(x))=y^2+2xy$

$Q(x-y,y)$ $\implies$ $f(xf(y))+f(yf(x-y))=2xy-y^2$
$Q(x-y,2y)$ $\implies$ $f((x+y)f(y))+f(yf(x-y))=2xy$
Subtracting, we get $f(xf(y)+yf(y))-f(xf(y))=y^2$

And so $f(xf(y)+yf(y))=f(xf(y))+f(yf(y))$ (remember that $P(y,0)$ $\implies$ $f(yf(y))=y^2$ )

And so $f(x+yf(y))=f(x)+f(yf(y))$ $\forall x,y$

And so $f(x+y)=f(x)+f(y)$ $\forall x\in\mathbb R,\forall y\in f^{[-1]}(\mathbb R^+\cup\{0\})$

But then $f(-x+y)=-f(x)+f(y)$ and so $f(x-y)=f(x)+f(-y)$
And so $f(x+y)=f(x)+f(y)$ $\forall x\in\mathbb R,\forall y\in f^{[-1]}(\mathbb R^-\cup\{0\})$
Q.E.D.

6) either $f(x)=x$ $\forall x$ , either $f(x)=-x$ $\forall x$
================================
Using $f(x+y)=f(x)+f(y)$ , $P(x,y)$ becomes $f(xf(x))+f(xf(y))=f(yf(x))+x^2$ and so $f(xf(y))=f(yf(x))$

$P(x+y,0)$ $\implies$ $f(xf(x))+f(xf(y))+f(yf(x))+f(yf(y))=(x+y)^2$ and so $f(xf(y))=xy$

Setting $y=f(1)$ and using $f(yf(y))=y^2$ , we get $f(x)=xf(1)$

Plugging this in original equation, we get $f(1)^2=1$
Hence the two solutions :

either $f(1)=1$ and so $\boxed{f(x)=x}$ $\forall x$

either $f(1)=-1$ and so $\boxed{f(x)=-x}$ $\forall x$

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jgnr

1343 posts

jgnr

#11 h Dec 21, 2010, 5:30 PM • 9 Y

Y by KMI, lmht, Abbas11235, myh2910, tiendung2006, kido2006, Adventure10, Mango247, and 1 other user

Another proof of injectivity: $f(x)=f(y)$ implies $x^2=f(xf(x))=f(xf(y))=f((y-x)f(x))+x^2$ , which gives $f((y-x)f(x))=0$ . So $y=x$ or $x=0$ , both of which implies $x=y$ .

Another way to continue from (3):
$f(xf(y))=f((y-x)f(x))+x^2=-f((x-y)f(x))+x^2=-f(yf(x-y))-(x-y)^2+x^2=f(yf(y-x))-(x-y)^2+x^2=f(-xf(y))+y^2-(x-y)^2+x^2=-f(xf(y))+2xy$
so $f(xf(y))=xy$ , and similarly we get $f(yf(x))=xy$ . Hence $f(xf(y))=f(yf(x))$ , which gives $xf(y)=yf(x)$ . So $f(x)=cx$ , and we easily get $c=\pm1$ .

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pco

23515 posts

pco

#12 h Dec 21, 2010, 5:43 PM • 2 Y

Y by Adventure10, Mango247

Johan Gunardi wrote:

Another proof of injectivity: $f(x)=f(y)$ implies $x^2=f(xf(x))=f(xf(y))=f((y-x)f(x))+x^2$ , which gives $f((y-x)f(x))=0$ . So $y=x$ or $x=0$ , both of which implies $x=y$ .

Another way to continue from (3):
$f(xf(y))=f((y-x)f(x))+x^2=-f((x-y)f(x))+x^2=-f(yf(x-y))-(x-y)^2+x^2=f(yf(y-x))-(x-y)^2+x^2=f(-xf(y))+y^2-(x-y)^2+x^2=-f(xf(y))+2xy$
so $f(xf(y))=xy$ , and similarly we get $f(yf(x))=xy$ . Hence $f(xf(y))=f(yf(x))$ , which gives $xf(y)=yf(x)$ . So $f(x)=cx$ , and we easily get $c=\pm1$ .

Yesssss, quite OK for both parts and quite nice !
Much quicker and simpler than my ugly steps. Congrats !

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Dijkschneier

131 posts

Dijkschneier

#13 h Jun 16, 2011, 5:36 PM • 1 Y

Y by Adventure10

No need for injectivity !
$P(0,y) \implies f(0)=f(yf(0))$ , and if f(0) is not zero, then f is constant and we get a contradiction. Hence f(0)=0.
$P(x,0) \implies f(xf(x))=x^2$ and $P(x,-x) \implies f(-xf(x))=-x^2$ and in particular, f is onto.
1) $f(x)=0 \iff x=0$
Let $c\in \mathbb{R}$ such that f(c)=0.
$P(c,0) \implies f(cf(c))=c^2 \implies 0=c^2 \implies c=0$
2) $f(d)=1 \implies d=\pm 1$
Let $d\in \mathbb{R}$ such that f(d)=1.
$P(d,0) \implies f(df(d))=d^2 \implies 1=d^2 \implies d=\pm 1$
3) $f(x)=y^2 \implies x=yf(y)$
Let x,y $\neq 0$ and define $s=\frac{x}{f(y)}$ .
$P(s,y-s) \implies f(x)=f((y-s)f(s))+s^2$
If f(x)=s², then f((y-s)f(s))=0, hence y-s=0 or f(s)=0, y=s or s=0, y=s, f(y)=f(s), x=sf(s).
As f is onto, s take all the real values when y changes in $\mathbb{R}$ , and hence the desired.
4) $f(2x)=2f(x)$
$P(x,x) \implies f(xf(2x))=f(xf(x))+x^2=2x^2=(\sqrt{2}x)^2 \\ \implies xf(2x)=\sqrt{2}xf(\sqrt{2}x) \\ \implies f(2x)=2f(x)$

Now from f(f(1))=1 and 2), we see that f(1)=1 or f(1)=-1.
1) First case : f(1)=1
$f(2x)=2f(x) \implies f(2)=2$
$P(1,y) \implies f(f(y+1))=f(y)+1$
$P(2,y) \implies f(2f(y+2))=f(2y)+4 \\ \implies f(f(y+2))=f(y)+2 \\ \implies f(y+1)+1=f(y)+2 \\ \implies f(y+1)=f(y)+1$
Now, $f(y)+1=f(f(y+1))=f(f(y)+1)=f(f(y))+1 \implies f(y)=f(f(y))$ and since f is onto we get f(x)=x for all real x.
2) Second case : f(1)=-1
$f(2x)=2f(x) \implies f(2)=-2$
$P(1,y) \implies f(f(y+1))=f(-y)+1$
$P(2,y) \implies f(2f(y+2))=f(-2y)+4 \\ \implies f(f(y+2))=f(-y)+2 \\ \implies f(-y-1)+1=f(-y)+2 \\ \implies f(-y-1)=f(-y)+1 \\ \implies f(y-1)=f(y)+1 \implies f(y+1)=f(y)-1$
Now, $f(-y)+1=f(f(y+1))=f(f(y)-1)=f(f(y))+1 \implies f(f(y))=f(-y)$ .
$f(x^2)=f(f(xf(x))=f(-xf(x))=-x^2$ and $f(-x^2)=f(f(-xf(x))=f(xf(x))=x^2$ so f(x)=-x for all real x.

Hence the two solutions : f(x)=x and f(x)=-x.

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JuanOrtiz

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JuanOrtiz

#14 h Jun 20, 2014, 1:17 AM • 2 Y

Y by Adventure10, Mango247

Sorry for not translating.

Sea la ecuación original $(1)$ .

Con $x=0$ en $(1)$ obtenemos que $f(0)=0$ ya que $f(yf(0))=f(0)$ y $f$ no puede ser constante.

Con $y=0$ en $(1)$ obtenemos $f(xf(x))=x^2$ , ecuación a la que llamaré $(2)$ , y llamemos $g(x)=xf(x)$ . Con $y=-x$ en $(1)$ obtenemos que $f(-xf(x))=-x^2$ , ecuación a la que llamaré $(3)$ . Con ésto, tenemos ya que $f$ es suryectiva.

Supongamos que existe $a \neq 0$ tal que $f(a)=0$ . Entonces en $(1)$ con $a=x$ tendremos que $a^2=f(af(a+y))$ para toda $y$ . Tomemos $e_0$ un número distinto de $a^2$ . Por suryectividad existe $e_1$ tal que $f(e_1)=e_0$ . Luego, existe $e_2$ tal que $f(e_2)=e_1/a$ . Con $y=e_2-a$ tendremos que $e_0=a^2$ , contradicción. Entonces $f(a)=0$ sii $a=0$ .

Ahora supongamos $f(a)=f(b)$ , y sea $k=f(a)$ . En $(1)$ con $x=a$ , $y=b-a$ tendremos, por $(2)$ , que $a^2=f(af(a))=f(af(b))=f(xf(x+y))=f(yf(x))+a^2$ y entonces $f(yf(x))=0$ . Por tanto, $yf(x)=0$ . Si $f(x)=0$ entonces $f(a)=f(b)=0$ entonces $a=b=0$ . De lo contrario, $y=0$ y así $a=b$ . Obtenemos que $f$ es inyectiva. Por tanto, $f$ es biyectiva.

Por $(2)$ , $f(g(x))=x^2$ y $f(-g(x))=-x^2$ . Con ésto es fácil ver que $g$ es suryectiva (dada la inyectividad de $f$ ). Sea $a$ un real cualquiera. Si $a=0$ tengo $f(0)=-f(-0)$ . Sea $b$ tal que $g(b)=a$ . Entonces vemos que $f(a)=-f(-a)$ . De aquí, $f$ es impar.

Ahora mostraré que $f(nxf(-x))=nx^2$ para todo $n \ge 0$ entero, usando inducción sobre $n$ . Para $n=0$ es obvio. Lo demostraré para $n$ usando que es cierto para $n-1$ . Tenemos por imparidad y por $(2)$ que $f((n-1)g(x))=(n-1)f(g(x))$ para todo $x$ . De aquí vemos que $f((n-1)x)=(n-1)f(x)$ dado que $g$ es suryectiva. Ahora, en $(1)$ sea $y=na$ y $x=-a$ para cualquier $a$ . Tendremos que

$-a^2(n-1)=f((n-1)af(-a))=f(-af((n-1)a))=f(xf(x+y))=a^2+f(nxf(-x))$

Y de aquí, $f(nxf(-x))=-nx^2$ para toda $a$ . La inducción está completa.

Entonces, usando el argumento de arriba, pues $g$ es suryectiva, encontramos que $f(na)=nf(a)$ para todo $n \ge 0$ entero. Por imparidad vemos fácilmente que $f(ra)=rf(a)$ para todo $r$ racional, para todo $a$ real. A ésta ecuación le llamo $(4)$ .

En $(2)$ con $x=1$ y con $x=f(1)$ obtenemos que $f(f(1))=1$ y luego que $f(f(1))=f(1)^2$ . Luego $f(1)=1$ ó $-1$ . Dado que $f$ es impar, si defino $f_1(x)=-f(x)$ , seguirá cumpliendo $(1)$ , y entonces puedo suponer sin pérdida de generalidad que $f(1)=1$ . De $(4)$ , se da $f(r)=r$ para $r$ racional.

De $(1)$ con $x=1$ observemos que $f(f(y+1))=f(y)+1$ . A ésto le llamo $(5)$ .

Ahora en $(1)$ sea $x \neq 0$ un racional cualquiera y usemos $(5)$ y $(4)$ . Resulta que

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

y así $f(x+y-1)+1=f(y)+x$ . Sea $r=x-1$ , se da que $f(r+y)=f(y)+r$ para todo $r \neq -1$ racional y para todo $y$ . A ésto le llamo $(6)$ .

Utilizemos $(5)$ y $(6)$ . Con $r=1$ tenemos que $f(x+1)=1+f(x)$ y entonces $f(x)=1+f(x-1)$ . Resulta que $f(f(x))=f(x-1)+1=f(x)$ para todo $x$ . Por inyectividad tendremos $f(x)=x$ para toda $x$ . Recordemos que supusimos sin pérdida de generalidad que $f(1)=1$ , entonces en realidad $-f$ también cumple.

Por ende, las únicas soluciones son:

$f(x)=x$

$f(x)=-x$

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Aiscrim

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Aiscrim

#15 h Sep 26, 2014, 8:55 AM • 2 Y

Y by Adventure10, Mango247

Solution

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ZETA_in_olympiad

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ZETA_in_olympiad

#45 h May 6, 2022, 9:15 PM

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Let $P(x,y)$ denote the given assertion. And let $P(x,y)\circ P(u,v)$ denote the combination of $P(x,y)$ and $P(u,v).$

Trivially we have $f(x)=0$ iff $x=0.$ We now prove a lemma.

Lemma: $f$ is injective.
Proof. If $f(m)=f(n)$ for some $m,n,$ then $m^2=f(mf(m))=f(mf(n))=m^2+f((n-m)f(m))\implies f((n-m)f(m))=0 \implies m=n.$ $\blacksquare$

$P(x,0)\circ P(-x,0)\implies f(xf(x))=f(-xf(-x))\implies f(-x)=-f(x).$
$P(x,x) \circ P(x,-2x)\implies f(2xf(x))=f(xf(2x))\implies f(2x)=2f(x).$
$P(1,x)\implies f(f(x+1))=f(x)+1.$
$P(1,0)\circ P(1,1)\implies f(1)=1\text{ or } f(1)=-1.$
$P(x,-x-1)\circ P(x+1,-1)\implies$
$f(x)= \begin{cases} x &\text{if }f(1)=1, \\ -x &\text{if }f(1)=-1,\end{cases}$ both of which fit.

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awesomeming327.

1721 posts

awesomeming327.

#47 h Aug 4, 2022, 8:39 PM

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:skull:

Let $P(x,y)$ denote the assertion. $P(0,y)$ gives $f(0)=f(yf(0)).$ If $f(0)$ is not zero, then $yf(0)$ is surjective, which means $f(x)$ is constant. However, back in the original equation, this implies $x^2=0$ so $f(x)$ is not constant, which means $f(0)=0.$ Now, $P(x,0)$ gives $f(xf(x))=x^2$ and $P(x,-x)$ gives $f(-xf(x))=-x^2.$

$~$
Suppose $f(x)=f(y)$ then $f(xf(x))=f(xf(y))=x^2$ for all $y.$ However, $f(xf(y))=f((y-x)f(x))+x^2$ which implies $f((y-x)f(x))=0.$ Note that if $f(x)=0$ then $0=f(xf(x))=x^2$ which implies $x=0$ , but this means $f$ is injective.

$~$
Now, $P(1,y)$ implies $f(f(1+y))=f(yf(1))+1$ , but $f(f(1))=1$ so $f(1)f(f(1))=f(1)$ which implies $f(f(1)f(f(1)))=f(f(1))=f(1)^2$ so $f(1)^2=1.$ Now, WLOG assume $f(1)=1.$ $P(x,x)$ gives $f(xf(2x))=2x^2=f(\sqrt{2}xf(\sqrt{2}x))$ so $f(2x)=\sqrt{2}f(\sqrt{2}x)$ so $f(2x)=2f(x).$ In particular, $f(2)=2$ and $P(2,x)$ implies $2f(f(2+y))=2f(y)+4$ which implies $f(y+2)=f(y)+2$

$~$
Since $f(f(1+y))=f(y)+1$ This implies $f(f(2+y))=f(y)+2=f(y+1)+1$ which implies $f(1+y)=f(y)+1$ so $f(1+y)=1+y$ so $f(y)=y.$ The other case gives $f(y)=-y$ and both work.

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Jalil_Huseynov

439 posts

Jalil_Huseynov

#48 h Dec 26, 2022, 2:17 PM • 2 Y

Y by Ibrahim_K, farhad.fritl

IMO SL 2009 A7 wrote:

Find all functions $f$ from the set of real numbers into the set of real numbers which satisfy for all $x$ , $y$ the identity $f\left(xf(x+y)\right) = f\left(yf(x)\right) +x^2$

Let $P(x,y)$ be assertion of the given equation. $P(0,y)\implies f(yf(0))=f(0)$ . If $f(0)\ne 0$ , then $f(x)\equiv c$ for all $x$ which doesn't work. So, $f(0)=0$ . $P(x,0) \implies f(xf(x))=x^2$ .
Assume there is $t\ne 0$ such that $f(t)=0$ . $P(t,0)\implies t^2=0$ which is contradiction. So, $f$ is injective at $0$ .
Assume there are $a\ne b$ such that $f(a)=f(b)$ . Obviously, $ab\ne 0$ . $P(a,b-a)\implies a^2=f(af(a))=f(af(b))=f((b-a)f(a))+a^2 \implies f((b-a)f(a))=0 \implies (b-a)f(a)=0$ . Since $a\ne 0,$ we get $a=b$ , which is contra. So, there are no such $a,b$ , which means $f$ is injective.
$f(xf(x))=x^2=(-x)^2=f(-xf(-x))\implies xf(x)=-xf(-x)\implies f(-x)=-f(x)$ .
$P(x,x)\implies f(xf(2x))=2x^2=(\sqrt2 x)^2=f(\sqrt2 xf(\sqrt2 x)) \implies xf(2x)= \sqrt2 xf(\sqrt2 x) \implies f(\sqrt2 z)=\sqrt2 f(z)\implies f(2x)=\sqrt2 f(\sqrt2 x)=2f(x)\implies f(2x)=2f(x)$ .
$P(2,2x-2)\implies f(2f(2x))=f((2x-2)f(2))+4\implies f(f(x))=f((x-1)f(1))+1$ . Let $Q(x)$ be assertion of $f(f(x))=f((x-1)f(1))+1$ .
$Q(2x)\implies f((2x-1)f(1))+1=f(f(2x))=2f(f(x))=2f((x-1)f(1))+2=f((2x-2)f(1))+2$ . Rewrite $(2x-1)f(1)=r \implies f(r+f(1))=f(r)+1$ for all $r\in\mathbb{R}$ . Now take $r=(s-1)f(1) \implies f(sf(1))=f((s-1)f(1))+1=f(f(s))\implies sf(1)=f(s) \implies f(x)=cx$ for all $x$ .
Since $f(f(1))=1$ , we get $c=1$ or $c=-1$ . So we get $2$ solutions $f(x)=x$ and $f(x)=-x$ and both indeed work.

This post has been edited 1 time. Last edited by Jalil_Huseynov, Dec 26, 2022, 2:21 PM

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EthanWYX2009

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EthanWYX2009

#49 h Jun 23, 2023, 3:27 PM

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Let $x=0,$ then $f(0)=f(yf(0)).$ If $f(0)\neq 0,$ then $f(y)\equiv C$ where ${C}$ is a constant $.$
Then $0=0+x^2,x\in\mathbb R,$ which is obviously wrong $.$ Therefore $f(0)=0.$
Let $y=0,$ then $f(xf(x))=x^2\cdots (*).$ We have $f(x)=0\Leftrightarrow x=0.$
If $\exists a,b\in\mathbb R,a\neq b,f(a)=f(b),$ then
$a^2\stackrel{(*)}{=}f(af(a))=f(af(b))=f((b-a)f(a))+a^2\Rightarrow f((b-a)f(a))=0$ $\Rightarrow (b-a)f(a)=0\Rightarrow f(a)=0\Rightarrow a=b=0$ which is not true $.$ Therefore for $\forall a,b\in\mathbb R,f(a)\neq f(b).$ As for $x\neq 0,$
$f(xf(x))\stackrel{(*)}{=}x^2=(-x)^2\stackrel{(*)}{=}f(-xf(-x))\Rightarrow xf(x)=-xf(-x)\Rightarrow f(x)=-f(-x).$ Let $x=y\neq 0$ , we have $f(xf(2x))=f(xf(x))+x^2\stackrel{(*)}{=}2x^2=(\sqrt 2x)^2\stackrel{(*)}{=}f(\sqrt 2xf(\sqrt 2x)).$ Then
$xf(2x)=\sqrt 2xf(\sqrt 2x)\Rightarrow f(2x)=\sqrt 2f(\sqrt 2x)\Rightarrow f(2x)=\sqrt 2f(\sqrt 2x)=2f(x).$ As $f(f(1))\stackrel{(*)}{=}1,f^2(1)\stackrel{(*)}{=}f(f(1)f(f(1)))=f(f(1))=1\Rightarrow f(1)=\pm 1.$
If $f(1)=1.$ Then $f(2)=2f(1)=2.$ Let $x=1,f(f(y+1))=f(yf(1))+1=f(y)+1.$ Let $x=2,$
$2f(f(y+2))=f(2f(y+2))=f(yf(2))+4=f(2y)+4=2f(y)+4\Rightarrow f(f(y+2))=f(y)+2.$ $\therefore f(y)+2=f(f(y+2))=f(y+1)+1\Rightarrow f(y+1)=f(y)+1=f(f(y+1))$ $\Rightarrow f(y+1)=y+1\Rightarrow f(x)=x,\forall x\in\mathbb R.$ If $f(1)=-1.$ Then $f(2)=2f(1)=-2.$ Let $x=1,f(f(y+1))=f(yf(1))+1=f(-y)+1.$ Let $x=2,$
$2f(f(y+2))=f(2f(y+2))=f(yf(2))+4=f(-2y)+4=2f(-y)+4\Rightarrow f(f(y+2))=f(-y)+2.$ $\therefore f(-y)+2=f(f(y+2))=f(-(y+1))+1\Rightarrow f(-y-1)=f(-y)+1=f(f(y+1))$ $\Rightarrow f(y+1)=-y-1\Rightarrow f(x)=-x,\forall x\in\mathbb R.$ Therefore $\boxed{f(x)=\pm x}.\blacksquare$

This post has been edited 2 times. Last edited by EthanWYX2009, Jun 24, 2023, 12:40 AM

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CyclicISLscelesTrapezoid

372 posts

CyclicISLscelesTrapezoid

#50 h Nov 3, 2023, 12:23 AM • 1 Y

Y by sabkx

Let $P(x,y)$ denote the assertion in the problem. The only solutions are $f(x) \equiv x$ and $f(x) \equiv -x$ , which work.

Claim 1: $f(x)=0$ if and only if $x=0$ .

Proof: If $f(0) \ne 0$ , then $P(0,x)$ gives $f(0)=f(xf(0))$ , so $f$ is constant, a contradiction. If $f(x)=0$ , then $P(x,0)$ gives $x^2=0$ , so $x=0$ . $\square$

Claim 2: $f$ is bijective.

Proof: $P(x,0)$ gives $f(xf(x))=x^2$ and $P(x,-x)$ gives $f(-xf(x))=-x^2$ , so $f$ is surjective. To prove that $f$ is injective, suppose that $f(a)=f(b)$ for real numbers $a$ and $b$ . Then, $P(a,b-a)$ implies $f(af(b))=f((b-a)f(a))+a^2$ . We also have $f(af(b))=f(af(a))=a^2$ , so $f((b-a)f(a))=0$ , which implies $a=0$ or $a=b$ by Claim 1. If $a=0$ , then Claim 1 implies $a=b$ , as desired. $\square$

Claim 3: $f$ is odd.

Proof: Plug in $P(x,0)$ and $P(-x,0)$ to get $f(xf(x))=f(-xf(-x))=x^2$ , which gives $f(-x)=-f(x)$ . $\square$

Claim 4: $f$ is additive.

Proof: By $P(x,\tfrac{y}{f(x)})$ and $P(x,-\tfrac{y}{f(x)}-2x))$ for nonzero $x$ , we have
$f(xf(x+\tfrac{y}{f(x)}))=f(y)+x^2=f((\tfrac{y}{f(x)}+2x)f(x))-x^2 \implies f(y+2xf(x))=f(y)+2x^2.$ Setting $y=0$ gives $f(2xf(x))=2x^2$ , so if $z=2xf(x)$ , then $f(y+z)=f(y)+f(z)$ . Replacing $y$ with $-y$ and using oddness gives $f(y-z)=f(y)+f(-z)$ . However, notice that
$\mathbb{R}=f^{-1}(\mathbb{R})=f^{-1}(\{2x^2 \colon x \in \mathbb{R}\}) \cup f^{-1}(\{-2x^2 \colon x \in \mathbb{R}\})=\{2xf(x) \colon x \in \mathbb{R}\} \cup \{-2xf(x) \colon x \in \mathbb{R}\},$ so the possible values of $z$ and $-z$ comprise all of $\mathbb{R}$ , which means $f$ is additive. $\square$

Thus, $P(x,y)$ gives
$f(yf(x))+x^2=f(xf(x)+xf(y))=f(xf(x))+f(xf(y)) \implies f(xf(y))=f(yf(x)) \implies xf(y)=yf(x),$ so $f(x) \equiv cx$ for some real number $c$ . We also have $f(xf(x))=x^2$ , so $c \in \{1,-1\}$ , as desired. $\square$

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amogususususus

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amogususususus

#51 h Mar 14, 2024, 9:39 AM

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Let $P(x,y)$ denote the assertion in the problem. The only solution that satisfies are $f(x)=x \ \forall \ x \in \mathbb{R}$ and $f(x)=-x \ \forall \ x \in \mathbb{R}$ .

Claim 1. $f(x)=0$ if and only if $x=0$ .
Proof. Note that $f$ is not constant, from $P(0,x)$ we have $f(0)=f(xf(0))$ .
If $f(0)\neq 0$ , then $xf(0)$ can be any value in $\mathbb{R}$ . Therefore $f$ is constant, contradiction. Hence $f(0)=0$ . If $f(x)=0$ for some $x \in \mathbb{R}$ , from $P(x,-x)$ we can derive $x^2=0$ , implying $x=0$ . Hence $f(x)=0$ if and only if $x=0$ .

Claim 2. $f(xf(x))=x^2 \ \forall \ x \in \mathbb{R}$ .
Proof. The proof comes immediately from $P(x,0)$ .

Claim 3. $f$ is injective.
Proof. If there exist $a,b \in \mathbb{R}$ such that $f(a)=f(b)$ , then from $P(b,a-b)$ we have
$f(bf(a))=f((a-b)f(b))+b^2$ From claim $2$ , we have $f(bf(a))=f(bf(b))=b^2$ . Therefore $f((a-b)f(b))=0$ .
From claim $1$ , we can conclude $(a-b)f(b)=0$ . If $f(b)=0$ then we have $a=b=0$ from claim $1$ . If $f(b)\neq 0$ , then $a-b=0$ giving $a=b$ . Hence $f$ is injective.

Claim 4. $f(2x)=2f(x) \ \forall \ x \in \mathbb{R}$ .
Proof. From $P(x,x)$ we have
$f(xf(2x))=f(xf(x))+x^2=2x^2$ From claim $2$ , we have $f(\sqrt{2}xf(\sqrt{2}x))=(\sqrt{2}x)^2=2x^2$ . Since $f$ is injective, we can conclude
$xf(2x)=\sqrt{2}xf(\sqrt{2}x)$ Obviously the claim is true for $x=0$ , if $x\neq 0$ then we have $f(2x)=\sqrt{2}f(\sqrt{2}x)$ . Substituting $x=\frac{\sqrt{2}}{2}x$ gives $f(\sqrt{2}x) =\sqrt{2}f(x)$ . Therefore,
$f(2x)=\sqrt{2}f(\sqrt{2}x)=2f(x)$ As desired.

From claim $2$ , we have $f(f(1))=1$ . Substituting $x=f(1)$ to claim $2$ gives $f(1)^2=1$ .

Case 1. $f(1)=1$
From claim $2$ we have $f(-f(-1))=1$ , since $f$ is injective we have $f(-1)=-1$ .
From $P(1,x+1)$ we have
$f(f(x+2))=f(x+1)+1$ From $P(2,x)$ and claim $4$ ,
$f(2f(x+2))=f(2x)+4 \Rightarrow 2f(f(x+2))=2f(x)+4 \Rightarrow f(f(x+2))=f(x)+2$ From those two equation we have
$f(x+1)=f(x)+1 \Rightarrow f(x+2)=f(x+1)+1$ Substituting this back to the first equation we have $f(f(x+2))=f(x+2)$ , since $f$ is injective we have $f(x)=x \ \forall \ x \in \mathbb{R}$ .

Case 2. $f(1)=-1$
Since this case is very similar to the former, I will leave it to the reader. This case gives us $f(x)=-x \ \forall \ x \in \mathbb{R}$ .

Having all cases exhausted, we are done.

This post has been edited 8 times. Last edited by amogususususus, Jul 5, 2024, 10:39 AM

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math_comb01

662 posts

math_comb01

#52 h Mar 22, 2024, 3:47 PM

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2009 A7 wrote:

Find all functions $f$ from the set of real numbers into the set of real numbers which satisfy for all $x$ , $y$ the identity $f\left(xf(x+y)\right) = f\left(yf(x)\right) +x^2$

Fun problem!

We claim that $f \equiv id,-id$
Claim 1: $f(0)=0$
Proof
Claim 2: $f$ is injective at $0$
Proof
Claim 3: $f$ is bijective
Proof
Claim 4: $f$ is odd
Proof
Claim 5: $f$ is solution, so is $-f$
Proof
Claim 6: $f(1)=1$
Proof
Claim 7: $f(2x)=2f(x)$
Proof
Claim 8 $f(y+2)=f(y)+2$
Proof
Claim 9: $f$ is idempotent.
Proof
Hence the only solutions are indeed $f \equiv x,-x$

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pie854

243 posts

pie854

#53 h Apr 13, 2024, 11:42 AM

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First A7!

If $f(0)\neq 0$ then $(x,y)=\left(0,\frac x{f(0)}\right)\Rightarrow f(x)=f(0)$ , so $f$ is a constant function but this is clearly impossible. It follows that $f(0)=0$ .

Now if we put $y=0$ we find that $f(xf(x))=x^2$ . Suppose $f(t)=0$ for some non zero real $t$ , then $t^2=f(tf(t))=0$ a contradiction. Thus $f(x)=0$ iff $x=0$ . Now we wish to show that $f$ is injective. So let us assume $f(a)=f(b)\neq 0$ , then $(x,y)=(b,a-b)\Rightarrow f((a-b)f(b))+b^2=f(bf(a))=f(bf(b))=b^2.$ So $f((a-b)f(b))=0$ , and it follows from our earlier observation that $(a-b)f(b)=0$ , thus $a=b$ as desired.

$(x,y)=(1,0)\Rightarrow f(f(1))=1$ and so $(x,y)=(f(1),0)\Rightarrow 1=f(1)^2$ . We will suppose WLOG that $f(1)=1$ . Now $(x,y)=(1,-1)$ also gives $f(-1)=-1$ . If we set $(x,y)$ to be $(1,y-1)$ and $(-1,-y+1)$ then we find $\boxed{f(f(y))=f(y-1)+1}=f(-f(-y)).$ So by injectivity, $f(y)=-f(-y)$ and so $f$ is odd.

For the final blow, $\begin{align*} & (x,y)=(1-x,x)\Rightarrow f(1-x)=f(xf(1-x))+(1-x)^2\Rightarrow f(x-1)=f(xf(x-1))-(x-1)^2 \\ & (x,y)=(x,-1)\Rightarrow f(xf(x-1))=f(-f(x))+x^2=-f(x-1)-1+x^2.\end{align*}$ Thus $f(x-1)=-f(x-1)-1+x^2-(x-1)^2$ , it follows that $f(x-1)=x-1$ . So $f(x)=x$ for all real $x$ . It is easy to check that this is a solution. If we had chosen $f(1)=-1$ then we would find $f(x)=-x$ which is also a solution.

This post has been edited 3 times. Last edited by pie854, Apr 13, 2024, 11:48 AM

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pokpokben

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pokpokben

#54 h Jul 3, 2024, 7:27 PM

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There is an alternative way to end the problem after after we proved the injectivity, $-f(x)=f(-x)$ and $f(2x)=2f(x)$
Put $P(x, x+y):$
$f(xf(2x+y))=f((x+y)f(x))+x^2;$ $P(2x,y):$
$f(2xf(2x+y))=f(yf(2x))+4x^2$ note that $f(yf(2x))+4x^2=2f(yf(x))+4x^2=2f((x+y)f(x))+2x^2$ $f(yf(x))+x^2=f((x+y)f(x))$ but LHS is equal to $f(xf(x+y))$
So due to injectivity $xf(x+y)=(x+y)f(x)$ , then $f(x)/x=cons$ , implying $f(x)=x$ or $f(x)=-x.$

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i3435

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i3435

#55 h Jul 4, 2024, 1:11 PM

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The answer is $x$ and $-x$ . Let $P(x,y)$ be the assertion.

$P(0,y)$ implies $f$ is constant or $f(0)=0$ , $f$ is not constant so $f(0)=0$ . $P(x,0)$ means $f(xf(x))=x^2$ . $f$ is injective at $0$ because if $c$ is such that $c=0$ , then $f(0)=c^2$ , so $c=0$ . If $f(x+y)=f(x)$ , $f(yf(x))+x^2=f(xf(x+y))=f(xf(x))=x^2$ , so $f(yf(x))=0$ and either $y=0$ or $f(x)=0$ , but when $f(x)=0$ then $x=0$ and $f(y)\neq 0$ unless $y=0$ . Thus $f$ is injective. In particular, since $f(xf(x))=x^2=f(-xf(-x))$ , then $xf(x)=-xf(-x)$ and $f(x)=-f(-x)$ .

$P(-x-y,x)$ gives, after taking negatives out, $f((x+y)f(y))=-f(xf(x+y))+(x+y)^2=-f(yf(x))+y(y+2x)$ . Plugging in $y=-2x$ and taking out negatives gives $f(xf(2x))=f(2xf(x))$ , so $xf(2x)=2xf(x)$ and $f(2x)=2f(x)$ . $P(x+y,x)$ gives $f((x+y)f(2x+y))=f(xf(x+y))+(x+y)^2=f(yf(x))+(x+y)^2+x^2$ . Thus if we let $a=y$ and $b=2x+y$ , $f\left(\frac{a+b}{2}f(b)\right)-f\left(\frac{a+b}{2}f(a)\right)=\frac{b^2-a^2}{2}$ for all $a,b\in \mathbb{R}$ . Thus $f((a+b)f(b))-f((a+b)f(a))=b^2-a^2$ . This means $f((x+y)f(y))-f((x+y)f(x))=y^2-x^2$ , but because of the equation $f((x+y)f(y))=(x+y)^2-x^2-f(yf(x))$ and its symmetric relation, $f(yf(x))=f(xf(y))$ , so $yf(x)=xf(y)$ . Thus $\frac{f(x)}{x}$ is constant for all $x$ , and we can get $x$ or $-x$ as the only solutions.

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vsamc

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vsamc

#56 h Jul 26, 2024, 7:36 PM • 2 Y

Y by ihatemath123, centslordm

Solution

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cosdealfa

27 posts

cosdealfa

#58 h Jan 11, 2025, 10:54 AM • 2 Y

Y by MR.1, Kaus_sgr

My first A7

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

Claim 1: $f(0)=0$
Proof: Suppose $f(0) \neq 0$
$P(0, \frac{1}{f(0)}): f(0)=f(1)$
$P(0, 1): f(0)=f(f(0))$
$P(1,0): f(f(1))=f(0)+1$
Plugging the first two relations into the third one, we get a contradiction. So $f(0)=0$ and moreover, if $f(x)=0$ for some $x \in \mathbb{R}$ then $x=0$ $\square$

Claim 2: $f$ is injective
Proof: Suppose we can find $a, b \in \mathbb{R}$ such that $a < b$ and $f(a)=f(b)$
$P(a, b-a): f(af(b))=f((b-a)f(a))+a^2$
But $f(af(b))=f(af(a))=a^2$ from $P(a,0)$ therefore $f((b-a)f(a))=0$ and using claim 1, this leads to $a=b$ , contradiction. So $f$ is injective. $\square$

Notice that $P(\sqrt{a}, 0)$ and $P(\sqrt{a},-\sqrt{a})$ give that $f$ is surjective.
Therefore we may find $c \in \mathbb{R}$ such that $f(c)=1$
$P(c,0): f(cf(c))=c^2$ $\Rightarrow c^2=1$ $\Rightarrow c \pm 1$
But it’s easy to see that if $f$ is a solution, so is $-f$ therefore we can assume WLOG that $c=1$ . So $f(1)=1$
Similarly, denote by $e$ the number for which $f(e)=-1$ .
By $P(e, -e): f(-ef(e))=-e^2$ $\Rightarrow e^2=1$ but from injectivity $e=-1$ therefore $f(-1)=-1$ .

Claim 3: $f$ is odd.
Proof:
$P(1, x):f(f(x+1))=f(x)+1$
$P(-1, x+1):f(-f(x))=f(-x-1)+1$
$P(1, -x-1): f(f(-x))=f(-x-1)+1$
$\Rightarrow f(-x)=-f(x)$ $\square$

Since $f$ is odd, $P(-1,x+1): -f(f(x))=-f(x+1)+1$
$\Rightarrow f(f(x))+1=f(x+1)=f(f(x)+1)$ so, by injectivity, $f(x)=x$ .
Therefore, the only solutions are $f(x)=x$ and $f(x)=-x$ $\blacksquare$

This post has been edited 2 times. Last edited by cosdealfa, Jan 11, 2025, 10:56 AM

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bin_sherlo

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bin_sherlo

#59 h Feb 26, 2025, 8:38 PM

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$f\left(xf(x+y)\right) = f\left(yf(x)\right) +x^2$ All such functions are $f(x)=x$ and $f(x)=-x$ . Let $P(x,y)$ be the assertion.

Claim: $f(0)=0$ .
Proof: $P(0,y)$ gives $f(0)=f(yf(0))$ and since $f$ is non-constant $f(0)=0$ .

Claim: $f$ is bijective.
Proof: $P(x,0)$ gives $f(xf(x))=x^2$ and $P(x,-x)$ implies $f(-xf(x))=-x^2$ hence $f$ is surjective. If $f(p)=0$ , then $0=f(pf(p))=p^2$ thus, $f$ is injective at $0$ . Suppose $f(a)=f(b)$ and $a\neq b$ . Plugging $y=a-x$ and $y=b-x$ gives $f((a-x)f(x))=f(xf(a))=f((b-x)f(x))$ but $x=a$ gives $f((b-a)f(a))=0$ which contradicts with $a\neq b$ .

Claim: $f$ is odd.
Proof: Since $f(xf(x))=x^2$ and $f$ is injective, $f(x)=-f(-x)$ .

Claim: There exists a constant such that $f(cx)=cf(x)$ .
Proof: Note that $x=y$ gives $f(xf(2x))=2x^2$ . Plugging $P(y,x)$ and $P(x,y-x)$ implies
$f(yf(x+y))-y^2=f(xf(y))=f(yf(x)-xf(x))+x^2\implies x^2+y^2=f(yf(x+y))-f(yf(x)-xf(x))$ Pick $y=2x$ to see that $f(x\sqrt3.f(2x\sqrt3))=6x^2=f(2xf(3x))$ and by injectivity $2f(3x)=\sqrt3f(2x\sqrt3)$ . Replace $x/3$ with $x$ in order to get $\frac{2}{\sqrt3}f(x)=f(\frac{2}{\sqrt3}x)$ so $c=\frac{2}{\sqrt3}$ satisfies.

Now, consider $P(cx,y)$ and divide both sides by $c$ which yields $f(xf(cx+y))=f(yf(x))+cx^2$ . By comparing this with $P(x,y)$ ,
$f(xf(cx+y))-f(xf(x+y))=(c-1)x^2$ Replace $y-x$ with $y$ to verify $f(xf((c-1)x+y))-f(xf(y))=(c-1)x^2$ . Let $c-1=d$ . We have
$f(xf(dx+y))=f(xf(y))+dx^2$ $x=1$ gives $f(f(y+d))=f(f(y))+d$ . Since $f$ is odd, comparing $x,y$ with $x,-y$ yields
$f(xf(dx+y))=f(xf(y))+dx^2$ $f(xf(dx-y))=-f(xf(y))+dx^2$ Sum these up to see $f(xf(dx+y))+f(xf(dx-y))=2dx^2$ . Now, pick $y=dx$ which gives $f(xf(2dx))=2dx^2=f(x\sqrt d.f(2x\sqrt d))$ hence $f(dx)=\sqrt d.f(x\sqrt d)$ or $f(x\sqrt d)=\sqrt d f(x)$ . Thus, $f(dx)=df(x)$ .
Since $f(f(y+d))=f(f(y))+d$ we have $f(f(dy+d))=f(f(dy))+d$ so $f(f(y+1))=f(f(y))+1 \ (\spadesuit)$ . Since $f(f(0))=0$ by induction we observe that $f(f(n))=n$ for all integers (note that $f$ is odd). Plugging $P(x,f(y))$ into the original equation implies $f(xf(x+f(y)))-x^2=f(f(x)f(y))$ and changing $x,y$ we see the following equality
$f(xf(x+f(y)))-x^2=f(yf(y+f(x)))-y^2$ Pick $x=-f(y)$ to get $y^2-f(y)^2=f(yf(y+f(-f(y))))=f(yf(y-f(f(y))))$ . Since $f(f(n))=n$ for integers, $y=n$ gives $f(n)^2=n^2$ so either $f(n)=n$ or $f(n)=-n$ .

Our first case is $f(1)=1$ , by $x=1$ we have $f(f(y+1))=f(y)+1$ and by $(\spadesuit)$ the equality
$f(f(y))+1=f(f(y+1))=f(y)+1$ holds and by injectivity $f(x)=x$ for all reals.

The latter case is $f(1)=-1$ , by $x=-1$ we have $f(f(y+1))=f(-y)+1$ and by $(\spadesuit)$ we get
$f(f(y))+1=f(f(y+1))=f(-y)+1$ By injectivity $f(x)=-x$ for all reals. We are done. $\blacksquare$

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Orzify

12 posts

Orzify

#61 h Mar 4, 2025, 5:15 PM

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I will present a very slick finish after showing injectivity and odd $f$ .

Claim: $f$ is injective and $f$ is odd.

standard

In conclusion, note that $P(x,-x-y)$ , $P(-x-y,y)$ and $P(y,x)$ give $-f(xf(y))=-f((x+y)f(x))+x^2$ $f((x+y)f(x))=-f(yf(x+y))+(x+y)^2$ $-f(yf(x+y))=-f(xf(y))-y^2$ Putting these together will give $f(xf(y))+x^2=-f(xf(y))-y^2+(x+y)^2\implies f(xf(y))=xy$ Henceforth by symmetry and injectivity $xf(y)=yf(x)$ and therefore $f(x)/x=c$ . Plugging this back into the original FE yields the solutions $f(x)=x$ and $f(x)=-x$

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ravengsd

17 posts

ravengsd

#62 h Mar 31, 2025, 1:09 PM • 1 Y

Y by Kaus_sgr

I hate using $2$ 's in FE's.
The only solutions are $f(x)=x$ and $f(x)=-x$ . These trivially work, so let's prove they are unique.
Denote $P(x,y)$ the given assertion.
Claim 1: $f(0)=0$
Proof: FTSoC suppose $f(0)\neq 0$
$P(0,\frac{1}{f(0)}): f(0)=f(1)$
$P(0,1):f(0)=f(f(0))$
$P(1,0): f(f(1))=f(f(0))=f(0)+1$ , which is indeed a contradiction with the above. Therefore $f(0)=0$ . $\square$
Claim 2: $f$ is surjective.
Proof: We will first prove $f$ takes every value between $0$ and $+\infty$ . Take $P(\sqrt{x},0)$ which gives: $x=f(\sqrt{x}f(\sqrt{x}))$ so $f$ takes every positive value. Now, for the values between $-\infty$ and $0$ , consider $P(-\sqrt{x},\sqrt{x})$ , which gives: $-x=f(\sqrt{x}f(-\sqrt{x}))$ , thus the claim is proved. $\square$
Claim 3: $f$ is injective.
Proof: Suppose FTSoC that there exist $a, b\in \mathbb{R}$ such that $a < b$ and $f(a)=f(b)$ .
$P(b-a,a):f((b-a)f(b))=f(af(b-a))+(b-a)^2$ so $f((b-a)f(b))>f(af(b-a))$ .
However, $P(a, b-a)$ yields $f(af(b-a))=f((b-a)f(a))+a^2$ , therefore $f(af(b-a))\geq f((b-a)f(a))=f((b-a)f(b))$ , contradiction.
So $f$ is injective. $\square$
Claim 4: $f$ is odd.
Proof:
$P(x, 0): f(xf(x))=x^2$
$P(-x,0):f(-xf(-x))=x^2$
By injectivity, we have $xf(x)=-xf(-x)$ so $f(x)=-f(x), \forall x \in \mathbb{R}$ $\square$

Let $c\in \mathbb{R}$ be such that $f(c)=1$ . $P(c,0): c^2=f(cf(c))=f(c)=1$ so $c=\pm 1$ .

Case 1: $f(1)=1$ . By claim 4, $f(-1)=-1$ .
$P(1,x-1):f(f(x))=f(x-1)+1$
In particular, $f(f(f(x)+1))=f(f(x))+1$
$P(-1,x+1):f(-f(x))=f(-x-1)+1 \Rightarrow f(f(x))+1=f(x+1)$
First, notice that $f(f(x))+1=f(f(f(x)+1))$ , so by claim 3, we have: $x+1=f(f(x)+1), \forall x\in \mathbb{R}$ . Setting $x\rightarrow 1$ , we get $f(2)=2$ .
Also, $f(f(x))=f(x-1)+1=f(x+1)-1$ , therefore $f(x)+2=f(x+2), \forall x\in \mathbb{R}$
$P(2,x-2):f(2f(x))=f(2(x-2))+4=f(2x)$ so by claim 3, $2f(x)=2x \Rightarrow f(x)=x, \forall x\in \mathbb{R}$ .

Case 2: $f(1)=-1$ can be treated similarly. It will imply $f(x)=-x, \forall x\in \mathbb{R}$ . $\blacksquare$

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