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

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3647 posts

orl

#1 h Nov 11, 2005, 5:28 PM • 12 Y

Y by nicky-glass, Pham_Quoc_Sang, nguyendangkhoa17112003, anantmudgal09, chessgocube, Adventure10, megarnie, ImSh95, mathmax12, Spiritpalm, and 2 other users

Let $\,{\mathbb{R}}\,$ denote the set of all real numbers. Find all functions $\,f: {\mathbb{R}}\rightarrow {\mathbb{R}}\,$ such that $f\left( x^{2}+f(y)\right) =y+\left( f(x)\right) ^{2}\hspace{0.2in}\text{for all}\,x,y\in \mathbb{R}.$

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e.lopes

349 posts

e.lopes

#2 h Nov 11, 2005, 5:46 PM • 7 Y

Y by nguyendangkhoa17112003, Aryan-23, chessgocube, Adventure10, ImSh95, Mango247, and 1 other user

The first step is to establish that $f(0) = 0$ . Putting $x = y = 0$ , and $f(0) = t$ , we get $f(t) = t^2$ .
Also, $f(x^2+t) = f(x)^2$ , and $f(f(x)) = x + t^2$ .
We now evaluate $f(t^2+f(1)^2)$ two ways.
First, it is $f(f(1)^2 + f(t)) = t + f(f(1))^2 = t + (1 + t^2)^2 = 1 + t + 2t^2 + t^4$ .
Second, it is $f(t^2 + f(1 + t)) = 1 + t + f(t)^2 = 1 + t + t^4$ . So $t = 0$ , as required.

It follows immediately that $f(f(x)) = x$ , and $f(x^2) = f(x)^2$ .
Given any $y$ , let $z = f(y)$ . Then $y = f(z)$ , so $f(x^2 + y) = z + f(x)^2 = f(y) + f(x)^2$ .
Now given any positive $x$ , take $z$ so that $x = z^2$ .
Then $f(x + y) = f(z^2 + y) = f(y) + f(z)^2 = f(y) + f(z^2) = f(x) + f(y)$ .
Putting $y = -x$ , we get $0 = f(0) = f(x + -x) = f(x) + f(-x)$ . Hence $f(-x) = - f(x)$ .
It follows that $f(x + y) = f(x) + f(y)$ and $f(x - y) = f(x) - f(y)$ hold for all $x$ , $y$ .

Take any $x$ .
Let $f(x) = y$ . If $y > x$ , then let $z = y - x. f(z) = f(y - x) = f(y) - f(x) = x - y = -z$ .
If $y < x$ , then let $z = x - y$ and $f(z) = f(x - y) = f(x) - f(y) = y - x$ .
In either case we get some $z > 0$ with $f(z) = -z < 0$ .
But now take $w$ so that $w^2 = z$ , then $f(z) = f(w^2) = f(w)^2 \ge 0$ .
Contradiction. So we must have $f(x) = x$

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probability1.01

2743 posts

probability1.01

#3 h Jul 26, 2006, 3:59 AM • 11 Y

Y by cobbler, sa2001, chessgocube, Adventure10, ImSh95, Mango247, and 5 other users

Set x = 0 to get $f(f(y)) = y+f(0)^{2}$ . We'll let $c = f(0)^{2}$ , so $f(f(y)) = y+c$ . Then

$f(a^{2}+f(f(b))) = f(b)+f(a)^{2}$
$f(a^{2}+b+c) = f(b)+f(a)^{2}$
$f(f(a^{2}+b+c)) = f(f(b)+f(a)^{2}) = b+f(f(a))^{2}$
$a^{2}+b+2c = b+(a+c)^{2}$
$2c = c^{2}+2ac$

Since this holds for all $a$ , it follows that $c = 0$ . Now we have $f(0) = 0 \implies f(f(y)) = y$ . Note that f must be surjective since we may let y vary among all reals, and f must be injective since if $f(a) = f(b)$ , then $a+f(x)^{2}= f(x^{2}+f(a)) = f(x^{2}+f(b)) = b+f(x)^{2}$ . Finally, if $u > v$ , then there is some $t$ s.t. $u = t^{2}+v$ , and so $f(u) = f(t^{2}+v) = f^{-1}(v)+f(t)^{2}> f^{-1}(v) = f(f(f^{-1}(v))) = f(v)$ . Hence f is strictly increasing. It is now clear that since $f(f(y)) = y$ , we must have $f(x) = x$ for all x.

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nhatthi

7 posts

nhatthi

#4 h Jul 26, 2006, 9:11 AM • 4 Y

Y by chessgocube, Adventure10, ImSh95, Mango247

$k=x_{1}+x_{2}+...+x_{n}$

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prester

119 posts

prester

#5 h Dec 27, 2009, 3:08 PM • 3 Y

Y by chessgocube, ImSh95, Adventure10

orl wrote:

Let $\,{\mathbb{R}}\,$ denote the set of all real numbers. Find all functions $\,f: {\mathbb{R}}\rightarrow {\mathbb{R}}\,$ such that
$f\left( x^{2} + f(y)\right) = y + \left( f(x)\right) ^{2}\hspace{0.2in}\text{for all}\,x,y\in \mathbb{R}.$

Denote with $P(x,y)$ the property $f(x^2+f(y))=y+f^{2}(x)$

Let $c=f^{2}(0)$ . So we have
$P(0,x) \Rightarrow f^{(2)}(x)=x+c\;\;\;x\in \mathbb{R} \: (*)$ where $f^{(2)}(x)=f(f(x)).$

Consider $a,b\in \mathbb{R}$ with $a\ne b$ and $f(a)=f(b)$ . So we have
$f^{(2)}(a)=f^{(2)}(b) \Rightarrow a+c=b+c \Rightarrow a=b.$ This is a contraddiction.
So it must be $f(a)\ne f(b)$ when $a \ne b$ and $f(x)$ must be injective.

Consider now $a\in \mathbb{R}$ . From $(*)$ we have $f^{(2)}(a-c)=a.$
Therefore, $\forall a\in \mathbb{R} \;\;\; \exists b\in \mathbb{R}, b=f(a-c),$ such that $f(b)=a.$
So $f(x)$ must be surjective.
Therefore $f(x)$ must be bijective.
Then there must be $u\in \mathbb{R}$ such that $f(u)=0$ and this point is unique.

So we have
$(1)$ $P(u,u) \Rightarrow f(u^2)=u$
$(2)$ $P(-u,u) \Rightarrow f(u^2)=u+f^{2}(-u)$

From $(1)$ and $(2)$ we have $u=u+f^{2}(-u) \Rightarrow f(-u)=0$

Since $f(x)$ must be bijective we have $u=0$ . Therefore $f(0)=0, \ c=0$

The equation $(*)$ becomes $f^{(2)}(x)=x\;\;\;\forall x\in \mathbb{R}\: (**)$

From $(**)$ we derive that $f^{(2)}(x)$ is strictly increasing.

So consider $a,b\in \mathbb{R}$ with $f(a)>f(b)$ . Then $f(f(a))>f(f(b))$ and this implies $a>b.$

Consider now $a,b\in \mathbb{R}$ with $f(a)<f(b)$ . Then $f(f(a))<f(f(b))$ and this implies $a<b.$

Therefore we derive that $f(x)$ is also strictly increasing.

Consider $a \in \mathbb{R}\;\;a\ne 0$ . If $f(a)>a$ then $f(f(a))>f(a) \Rightarrow a>f(a).$
This is a contraddiction.

If $f(a)<a$ then $f(f(a))<f(a) \Rightarrow a<f(a).$
This is again a contraddiction.

The last two results say that $\forall x\in \mathbb{R}, \;\;x \ne 0,$ it must be $f(x)=x.$

Since $f(0)=0$ we have that $\forall x \in \mathbb{R}$ it must be $f(x)=x.$

Therefore
$\boxed{f(x)=x,\ \forall x \in \mathbb{R}}$ is a solution of the initial equation and it is the unique solution.

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prester

119 posts

prester

#6 h Dec 29, 2009, 7:02 AM • 4 Y

Y by chessgocube, ImSh95, Adventure10, Mango247

prester wrote:

So consider $a,b\in \mathbb{R}$ with $f(a) > f(b)$ . Then $f(f(a)) > f(f(b))$ and this implies $a > b.$

Consider now $a,b\in \mathbb{R}$ with $f(a) < f(b)$ . Then $f(f(a)) < f(f(b))$ and this implies $a < b.$

Therefore we derive that $f(x)$ is also strictly increasing.

The above steps are obviously wrong.

We cannot derive that $f(f(a)) > f(f(b))$ because we do not know that $a > b$ but we do know that $f(a) > f(b)$ and then the hypothesys uses the assert we want to demonstrate. It is non-sense.

To demonstrate that $f(x)$ is strictly increasing, I did not find a best way than previous post (by probability1.01).
Let $a > b\;\;\; a,b \in \mathbb{R}.$ So $\exists u \in \mathbb{R}$ such that $a = u^2 + b$ .
So $f(a) = f(u^2 + b) = f^2(u) + f^{ - 1}(b) > f^{ - 1}(b) = f^{(2)}(f^{ - 1}(b)) = f(b)$ and therefore $f(x)$ is strictly increasing.

Then the demonstration can proceed as before.

Sorry for mistake.

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arshakus

769 posts

arshakus

#7 h May 12, 2010, 12:23 PM • 3 Y

Y by ImSh95, Adventure10, Mango247

probability1.01 wrote:

Set x = 0 to get $f(f(y)) = y+f(0)^{2}$ . We'll let $c = f(0)^{2}$ , so $f(f(y)) = y+c$ . Then

$f(a^{2}+f(f(b))) = f(b)+f(a)^{2}$
$f(a^{2}+b+c) = f(b)+f(a)^{2}$
$f(f(a^{2}+b+c)) = f(f(b)+f(a)^{2}) = b+f(f(a))^{2}$
$a^{2}+b+2c = b+(a+c)^{2}$
$2c = c^{2}+2ac$

Since this holds for all $a$ , it follows that $c = 0$ . Now we have $f(0) = 0 \implies f(f(y)) = y$ . Note that f must be surjective since we may let y vary among all reals, and f must be injective since if $f(a) = f(b)$ , then $a+f(x)^{2}= f(x^{2}+f(a)) = f(x^{2}+f(b)) = b+f(x)^{2}$ . Finally, if $u > v$ , then there is some $t$ s.t. $u = t^{2}+v$ , and so $f(u) = f(t^{2}+v) = f^{-1}(v)+f(t)^{2}> f^{-1}(v) = f(f(f^{-1}(v))) = f(v)$ . Hence f is strictly increasing. It is now clear that since $f(f(y)) = y$ , we must have $f(x) = x$ for all x.

heloo,
can you explain me how you get $f(f(a^{2}+b+c)) = f(f(b)+f(a)^{2}) = b+f(f(a))^{2}$ ??

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R.Maths

189 posts

R.Maths

#8 h May 12, 2010, 11:18 PM • 4 Y

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

I think it's easy to chek that $f(0)=0$

thereby $f(f(x))=x$

So , consider the sequence defined by : $f(f(f(....)))=x_{n}$ and $x_{0}=x$

So $f(f(x))=x \implies x_{n+2}=x_{n}$ so for all $n \in \mathbb N$ there existe

$a;b \in \mathbb R$ such that $x_{n}=a+b(-1)^{n}$

for $n=1$ we have $f(x)=a-b=a+b-2b=x-2b$ because $a+b=x$

replacing this values in the equation $f(f(x))=x$ and we will merely find that $b=0$

then $( \forall x \in \mathbb R ) f(x)=x$

Wich is indeed a solution for the initial fonctioanl equation .

Redwane.

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arshakus

769 posts

arshakus

#9 h May 13, 2010, 2:13 PM • 2 Y

Y by ImSh95, Adventure10

R.Maths wrote:

I think it's easy to chek that $f(0)=0$

thereby $f(f(x))=x$

So , consider the sequence defined by : $f(f(f(....)))=x_{n}$ and $x_{0}=x$

So $f(f(x))=x \implies x_{n+2}=x_{n}$ so for all $n \in \mathbb N$ there existe

$a;b \in \mathbb R$ such that $x_{n}=a+b(-1)^{n}$

for $n=1$ we have $f(x)=a-b=a+b-2b=x-2b$ because $a+b=x$

replacing this values in the equation $f(f(x))=x$ and we will merely find that $b=0$

then $( \forall x \in \mathbb R ) f(x)=x$

Wich is indeed a solution for the initial fonctioanl equation .

Redwane.

helooo,
van you explian me why So f(f(x))=x \implies x_{n+2}=x_{n} so for all n \in \mathbb N there existe???

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arshakus

769 posts

arshakus

#10 h May 13, 2010, 2:29 PM • 3 Y

Y by ImSh95, Adventure10, Mango247

e.lopes wrote:

The first step is to establish that $f(0) = 0$ . Putting $x = y = 0$ , and $f(0) = t$ , we get $f(t) = t^2$ .
Also, $f(x^2+t) = f(x)^2$ , and $f(f(x)) = x + t^2$ .
We now evaluate $f(t^2+f(1)^2)$ two ways.
First, it is $f(f(1)^2 + f(t)) = t + f(f(1))^2 = t + (1 + t^2)^2 = 1 + t + 2t^2 + t^4$ .
Second, it is $f(t^2 + f(1 + t)) = 1 + t + f(t)^2 = 1 + t + t^4$ . So $t = 0$ , as required.

It follows immediately that $f(f(x)) = x$ , and $f(x^2) = f(x)^2$ .
Given any $y$ , let $z = f(y)$ . Then $y = f(z)$ , so $f(x^2 + y) = z + f(x)^2 = f(y) + f(x)^2$ .
Now given any positive $x$ , take $z$ so that $x = z^2$ .
Then $f(x + y) = f(z^2 + y) = f(y) + f(z)^2 = f(y) + f(z^2) = f(x) + f(y)$ .
Putting $y = -x$ , we get $0 = f(0) = f(x + -x) = f(x) + f(-x)$ . Hence $f(-x) = - f(x)$ .
It follows that $f(x + y) = f(x) + f(y)$ and $f(x - y) = f(x) - f(y)$ hold for all $x$ , $y$ .

Take any $x$ .
Let $f(x) = y$ . If $y > x$ , then let $z = y - x. f(z) = f(y - x) = f(y) - f(x) = x - y = -z$ .
If $y < x$ , then let $z = x - y$ and $f(z) = f(x - y) = f(x) - f(y) = y - x$ .
In either case we get some $z > 0$ with $f(z) = -z < 0$ .
But now take $w$ so that $w^2 = z$ , then $f(z) = f(w^2) = f(w)^2 \ge 0$ .
Contradiction. So we must have $f(x) = x$

helooo,
can you explain me why Second, it is $f(t^2 + f(1 + t)) = 1 + t + f(t)^2 = 1 + t + t^4$

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arshakus

769 posts

arshakus

#11 h May 15, 2010, 9:59 AM • 3 Y

Y by ImSh95, Adventure10, Mango247

arshakus wrote:

e.lopes wrote:

The first step is to establish that $f(0) = 0$ . Putting $x = y = 0$ , and $f(0) = t$ , we get $f(t) = t^2$ .
Also, $f(x^2+t) = f(x)^2$ , and $f(f(x)) = x + t^2$ .
We now evaluate $f(t^2+f(1)^2)$ two ways.
First, it is $f(f(1)^2 + f(t)) = t + f(f(1))^2 = t + (1 + t^2)^2 = 1 + t + 2t^2 + t^4$ .
Second, it is $f(t^2 + f(1 + t)) = 1 + t + f(t)^2 = 1 + t + t^4$ . So $t = 0$ , as required.

It follows immediately that $f(f(x)) = x$ , and $f(x^2) = f(x)^2$ .
Given any $y$ , let $z = f(y)$ . Then $y = f(z)$ , so $f(x^2 + y) = z + f(x)^2 = f(y) + f(x)^2$ .
Now given any positive $x$ , take $z$ so that $x = z^2$ .
Then $f(x + y) = f(z^2 + y) = f(y) + f(z)^2 = f(y) + f(z^2) = f(x) + f(y)$ .
Putting $y = -x$ , we get $0 = f(0) = f(x + -x) = f(x) + f(-x)$ . Hence $f(-x) = - f(x)$ .
It follows that $f(x + y) = f(x) + f(y)$ and $f(x - y) = f(x) - f(y)$ hold for all $x$ , $y$ .

Take any $x$ .
Let $f(x) = y$ . If $y > x$ , then let $z = y - x. f(z) = f(y - x) = f(y) - f(x) = x - y = -z$ .
If $y < x$ , then let $z = x - y$ and $f(z) = f(x - y) = f(x) - f(y) = y - x$ .
In either case we get some $z > 0$ with $f(z) = -z < 0$ .
But now take $w$ so that $w^2 = z$ , then $f(z) = f(w^2) = f(w)^2 \ge 0$ .
Contradiction. So we must have $f(x) = x$

helooo,
can you explain me why Second, it is $f(t^2 + f(1 + t)) = 1 + t + f(t)^2 = 1 + t + t^4$

yes yes, I have just understand it

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DylanN

194 posts

DylanN

#12 h May 15, 2010, 12:58 PM • 4 Y

Y by shon804, ImSh95, Adventure10, Mango247

Here's my solution:

Taking $y = -f(x)^2$ we get $f(x^2 + f(y)) = 0$ i.e., $\exists b$ such that $f(b)=0$
Then $f(b^2+f(y))=y$ and so $f$ is bijective.

Then $f(b)^2= f(b^2+f(0))=f((-b)^2 +f(0)) = f(-b)^2$ so $f(-b)^2 = 0$
Hence $f(-b)=0=f(b)$ and so $-b=b$ since $f$ is injective, giving us $b=0$
So $f(0)=0$

Then $f(f(x))=x$
Suppose $x>y$ for some $x$ and $y$ . Let $x=y+z^2$
Then $f(x)=f(z^2+f(f(y)))=f(y)+f(z)^2>f(y)$ so $f$ is strictly increasing.

It is then well known that the only increasing solutions to $f(f(x))=x$ is $f(x)=x$

So $f(x)=x$ which works.

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arshakus

769 posts

arshakus

#13 h May 17, 2010, 11:32 AM • 3 Y

Y by ImSh95, Adventure10, Mango247

hi guys,
I found this solution in shortlist, but I think that it is not full, can you look at this solution and explain me dear reader!!!

It is easy to see that f is injective and surjective. From $f(x^2 + f(y)) =$
$f((-x)^2 +f(y))$ it follows that $f(x)^2 = (f(-x))^2$ , which implies $f(-x) = -f(x)$ because f is injective. Furthermore, there exists $z∈R$ such that
$f(z) = 0.$ From $f(-z) = -f(z) = 0$ we deduce that $z = 0$ . Now we
have $f(x^2) = f(x^2 + f(0)) = 0 + (f(x))^2 = f(x)^2$ , and consequently
$f(x) = f(sqrt(x)^2) > 0$ for all $x > 0$ . It also follows that $f(x) < 0$ for $x < 0$ .
In other words, f preserves sign.
Now setting $x > 0$ and $y = -f(x)$ in the given functional equation we
obtain
$f(x - f(x)) = f(sqrt(x)^2+ f(-x)) = -x + f(√x)2 = -(x - f(x)).$
But since f preserves sign, this implies that $f(x) = x for x > 0.$ Moreover,
since $f(-x) = -f(x),$ it follows that $f(x) = x$ for all x. It is easily verified
that this is indeed a solution.

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Rijul saini

904 posts

Rijul saini

#14 h Feb 26, 2011, 11:57 PM • 5 Y

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

orl wrote:

Let $\,{\mathbb{R}}\,$ denote the set of all real numbers. Find all functions $\,f: {\mathbb{R}}\rightarrow {\mathbb{R}}\,$ such that
$f\left( x^{2} + f(y)\right) = y + \left( f(x)\right) ^{2}\hspace{0.2in}\text{for all}\,x,y\in \mathbb{R}.$

Denote by $P(x,y)$ the assertion $f(x^2+f(y))=y+f^{2}(x)$ .

[hide="1) $f(0) = 0$ "] $P(0,0) \implies f(f(0)) = f(0)^2 \\ \\ P(0,x) \implies f(f(x)) = x+f(0)^2 \implies f(x) + f(0)^2 = f(f(f(x))) = f(x+f(0)^2)$

$P(x,0) \implies f(x^2+f(0)) = f(x)^2 \\ \\ P(-x,0) \implies f(x^2+f(0)) = f(-x)^2 \\ \\ \implies f(-x) = \pm f(x)$
But,
$f(-x) = f(x) \implies x+f(0)^2 = f(f(x)) = f(f(-x)) = -x+f(0)^2$
But this is true only for $x=0$ .

Now, take any non-zero real $a$ .
We have,
$-a +f(0)^2 = f(f(-a)) = f(-f(a)) = -f(f(a)) = -(a+f(0)^2) \\ \\ \implies f(0) = 0$ [/hide]

2) Sign of [math=inline]$f(x)$[/math] is the same as the sign of [aopsnowrap][math=inline]$x$[/math].[/aopsnowrap]

[hide="3) $f(x+1) = f(x) +1$ "]
We have,
$P(1,0) \implies f(1) = f(1)^2 \implies f(1) = 0 \ \text{or } f(1) = 1.$
But, if $f(1) =0$ , then
$P(x,1) \implies f(x^2) = 1+f(x)^2$
This is clearly not possible, because
$P(x,0) \implies f(x^2) = f(x)^2 \not = 1+f(x)^2$
Therefore, $f(1) =1$ .
Now,
$P(1,x) \implies f(f(x) + 1) = x+1 \implies f(x+1) = f(f(f(x) +1)) =f(x) +1$
since $f(f(x)) =x$ .[/hide]

[hide="4) $f(x)=x \ \forall x \in \mathbb{R}$ "]
We have,
$P(x,1-x^2) \implies f(x^2+ f(1-x^2)) = 1 + (f(x)^2 - x^2)$
Now,
$f(1-x^2) = 1+f(-x^2) = 1-f(x^2) \implies f(x^2+ f(1-x^2)) = f(1+(x^2 - f(x^2))) = 1+ f(x^2 - f(x^2))$
Also, using $f(x)^2 = f(x^2)$ , which has been proved earlier, we have
$1+ f(x^2 - f(x^2)) = f(x^2+ f(1-x^2)) = 1 + (f(x)^2 - x^2) = 1 + (f(x^2) - x^2)$
Therefore, we have,
$f(x^2 - f(x^2)) = f(x^2) - x^2$
Therefore, from 2nd lemma,
$x^2 - f(x^2) = 0 \iff f(y) = y\ \forall y \in \mathbb{R}^+$
Combining this with $f(-x) = -f(x)$ , we get the desired.[/hide]

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SnowEverywhere

801 posts

SnowEverywhere

#15 h Aug 7, 2011, 8:17 PM • 4 Y

Y by trying_to_solve_br, ImSh95, Adventure10, Mango247

Let $P(x,y)$ be the assertion that $f(x^2 + f(y))=y+f(x)^2$ . By $P(0,y)$ , we have that $f(f(y))=y+f(0)^2$ which implies that $f$ is a surjective function. Furthermore, if $f(x)=f(y)$ , then it follows that $x+f(0)^2=f(f(x))=f(f(y))=y+f(0)^2$ which implies that $x=y$ and hence that $f$ is injective. Now let $a \in \mathbb{R}$ be such that $f(a)=0$ . By $P(a,f(y))$ , it follows that $f(a^2+f(f(y)))=f(y)+f(a)^2=f(y)$ which implies that $y=f(f(y))+a^2$ since $f$ is injective. However, as previously established, $f(f(y))=y+f(0)^2$ which implies that $a^2 + f(0)^2=0$ and hence that $a=f(0)=0$ and $f(f(y))=y$ . Now by $P(x,0)$ , it follows that $f(x^2)=f(x)^2$ and furthermore that $f(x)^2=f(x^2)=f((-x)^2)=f(-x)^2$ which implies that $f(-x)=-f(x)$ since $f$ is injective. Now by $P(x,f(y))$ , it follows that $f(x^2+y)=f(x^2+f(f(y)))=f(y)+f(x)^2=f(y)+f(x^2)$ . Combining this with the fact that $f(-x)=-f(x)$ yields that $f(x+y)=f(x)+f(y)$ for all $x,y \in \mathbb{R}$ . Now note that for all $a \ge 0$ , $f(a)=f((\sqrt{a})^2)=f(\sqrt{a})^2 \ge 0$ . Hence $f$ is bounded on an interval which implies by the Cauchy functional equation that $f(x)=cx$ for some constant $c \in \mathbb{R}$ . Substituting this into the equation yields that $f(x)=x$ .

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SerdarBozdag

892 posts

SerdarBozdag

#43 h Sep 5, 2021, 10:47 AM • 1 Y

Y by ImSh95

$f$ is bijective. $P(0,y) \implies f(f(y))=y+f(0)^2$ .

$f(a)=0, P(x,y-f(x)^2) \implies f(f(y)-f(x)^2)=y-x^2 \implies f(f(y))=y-a^2$ and $a-a^2=f(0)$ .

$-a^2=f(0)^2=(a-a^2)^2 \implies a=0$ or $a^2-2a+2 \implies a=0 \implies f(f(x))=x$ . $P(x,0) \implies f(x^2)=f(x)^2 \implies f(x^2+f(y))=y+f(x^2) \implies f(x^2+y)=f(x^2)+f(y)$ for $y \rightarrow f(y)$ . $f(x^2)=f(x)^2 \implies f(x)=\pm f(-x)$ .

If $f(c)=f(-c), P(x,c)$ and $P(x,-c)$ together implies $c=0$ . Thus $f(x)=-f(-x)$ and $f(x^2+y)=f(x^2)+f(y) \implies f(-x^2+y)=-f(x^2-y)=-(f(x^2)+f(-y))=f(-x^2)+f(y) \implies f(x+y)=f(x)+f(y)$ for all $x,y \in R$ .

$f(x)^2+2f(x)f(y)+f(y)^2=(f(x)+f(y))^2=f(x+y)^2=f(x^2+2xy+y^2)=f(x^2)+f(2xy)+f(y^2) \implies 2f(x)f(y)=f(2xy) \implies 2f(x)=f(2x) \implies f(x)f(y)=f(xy)$ for $x,y \in R$ . With $f(x)+f(y)=f(x+y)$ we conclude $f(x)=x$ .

This post has been edited 1 time. Last edited by SerdarBozdag, Sep 5, 2021, 10:48 AM

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Marinchoo

407 posts

Marinchoo

#44 h Nov 14, 2021, 9:49 AM • 1 Y

Y by ImSh95

We'll use $P(x,y)$ for the assertion $(x,y)$ into the FE:

$P(0,y)$ gives $f(f(y))=y+f(0)^2\Longrightarrow f$ is bijective (as $y$ runs from $-\infty$ to $+\infty$ , $y+f(0)^2$ covers $\mathbb{R}$ , so $f$ in surjective and if we assume that $\exists a\neq b\in\mathbb{R}$ such that $f(a)=f(b)$ then $b+f(0)^2=f(f(b))=f(f(a))=a+f(0)^2$ , contradiction $\implies$ $f$ is bijective)

Therefore $\exists$ a unique $z\in\mathbb{R}$ such that $f(z)=0$ since $f$ is bijective:
$P(0,0)\Longrightarrow f(f(0))=f(0)^2$ $(1)$
$P(z,f(0))\Longrightarrow f(z^2+f(f(0)))=f(0)+f(z)^2\overset{(1)}{\Longrightarrow} f(z^2+f(0)^2)= f(0)\Longrightarrow z^2+f(0)^2=0\Longrightarrow z=0, \boxed{f(0)=0}$

Now $P(0,y)$ gives $f(f(y))=y$ and $P(x,0)$ gives $f(x^2)=f(x)^2\implies f(x)=-f(-x)$ . Using these two results we have that $P(x,f(y)):\quad f(x^2+y)=f(y)+f(x^2)$ . This is the Cauchy equation and as $f(x^2)=f(x)^2\geq 0$ this means that $f$ is monotonic, which implies that $f(x)=cx$ for some constant $c\in \mathbb{R}$ . As $c\times 1^2=f(1^2)=f(1)^2=c^2\times 1^2\Longrightarrow c\in\{0,1\}$ . The function $f\equiv 0$ is not a solution as $f$ is bijective, so the only possible solution is $f(x)=x$ . And indeed:
$f(x^2+f(y))=f(x^2+y)=x^2+y=y+f(x)^2$ Therefore $f(x)=x$ is the only solution to the FE.

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laikhanhhoang_3011

637 posts

laikhanhhoang_3011

#45 h Nov 15, 2021, 8:26 AM • 1 Y

Y by ImSh95

orl wrote:

Let $\,{\mathbb{R}}\,$ denote the set of all real numbers. Find all functions $\,f: {\mathbb{R}}\rightarrow {\mathbb{R}}\,$ such that $f\left( x^{2}+f(y)\right) =y+\left( f(x)\right) ^{2}\hspace{0.2in}\text{for all}\,x,y\in \mathbb{R}.$

$P(0;y)$ we have $f(f(y))=y+(f(0))^2, (1)$ so f is bijective
Exist a such that $f(a)=0$
$P(a;0):$ $f(a^2+f(0))=0=f(a)$ -> $a^2+f(0)=a, (2)$
Let $y=a$ in $(1):$ $f(0)=a+f(0)^2, (3)$
From $(2)$ and $(3)$ we have $f(0)=a=0$
We get $f(x^2)=f(x)^2, (*)$ and $f(f(y))=y$
$P(x;f(y)):$ $f(x+y)=f(x)+f(y), (4)$ for all $x \geq 0$ and all real $y$
$Q(x;y)$ in $(4)$
Let $x>0$ then from $(4)$ we have $0=f(0)=f(x)+f(-x)$ so f is odd
From $(*)$ we have $f(x) \geq 0$ for all $x \geq 0, (5)$
From $(4)$ and $(5)$ we have $f(x)=x$ , which is a familiar problem
Q.E.D

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iamnotAsif

19 posts

iamnotAsif

#46 h Dec 19, 2021, 3:57 PM • 1 Y

Y by ImSh95

A rather lengthy solution IG ;-;(Pls correct if I am wrong at some point)
We claim that only $f(x)=x$ which can be easily checked.

Let the assertion be $P(x,y)$ .
It is easy to see $f(x)$ is injective.
$P(x,-f(x)^2) \Rightarrow f(x^2+f(-f(x)^2)=0$ Since $f(x)$ is injective it follows that $x^2+f(-f(x)^2)=t ...(1)$ for some unique $t$
Now, Plugging $x=t$ in $(1)$ we get $t^2+f(0)=t$
Again, $P(x,t) \Rightarrow f(x^2)=f(x)^2 +t \Rightarrow f(0)=f(0)^2+t$
Combining the above two expressions we get $f(0)=0$
Plugging $x=f(x)$ in $(1)$ we get $f(-x)=-f(x)$
Again, $P(x,-x^2) \Rightarrow f(x^2-f(x)^2)=f(x)^2-x^2$
Now lets assume $f(a)=-a$ for some $a \neq 0$ also assume WLOG that $a>0$ since $f(-x)=-f(x)$
Then $P(\sqrt{a},a) \Rightarrow f(\sqrt{a})^2=-a$ which is a contradiction. So, $a=0 \Rightarrow f(x)^2=x^2$
Now we fix the point wise error.
assume for $f(a)=a,f(b)=-b$
$P(a,b) \Rightarrow f(a^2 -b)=a^2+b$ Contradiction.
$f(x)=x$ is the only solution(since $f(x)=-x$ don't work).

This post has been edited 3 times. Last edited by iamnotAsif, Dec 19, 2021, 4:03 PM

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megarnie

5538 posts

megarnie

#47 h Feb 17, 2022, 1:09 AM • 1 Y

Y by ImSh95

$P(0,x): f(f(x))=x+f(0)^2$ , so $f$ is bijective.

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

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

Note that $f(f(x)^2+f(y))=x^2+y+2f(0)^2$ . Suppose FTSOC that $f(0)\ne 0$ .

So we get $\begin{align*} x^2+y+2f(0)^2=x^2+2xf(0)^2+f(0)^4+y \\ 2f(0)^2=2xf(0)^2+f(0)^4 \\ 2=2x+f(0)^2\forall x \\ \end{align*}$
This is a contradiction because it implies that $f(0)^2+2\cdot 332.5=f(0)^2+665=2\implies f(0)^2=-663$ and also $f(0)^2+2\cdot 93.5=f(0)^2+187=2\implies f(0)^2=-185$ .

So $f(0)=0$ , which implies $f(f(x))=x$ .

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

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

So $f(x^2+y)=f(x^2)+f(y)$ .

Claim: $f$ is odd.
Proof: We have $f(x^2)=f(-x)^2=f(x)^2$ . If $f(x)=f(-x)$ for some $x$ , then we have $f(f(x))=f(f(-x))\implies x=-x\implies x=0$ . So $f(x)=-f(-x)$ for all $x$ since $f(0)=0$ . $\blacksquare$

For all reals $x$ and $y$ that are not both negative, we get $f(x+y)=f(x)+f(y)$ . Now if $u$ and $v$ are negative, we get $f((-u)+(-v))=-f(u+v)=f(-u)+f(-v)=-(f(u)+f(v))$ , so $f(u+v)=f(u)+f(v)$ . Thus, $f$ is additive.

From $f(x^2)=f(x)^2$ , $f$ is bounded over the interval (always nonegative) $[0,\infty]$ , so $f(x)=kx$ for some constant $k$ .

From involution we get $k=1$ or $k=-1$ . From $f(x^2)=f(x)^2$ , we get $k=1$ . So $\boxed{f(x)=x}$ is the only solution.

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ZETA_in_olympiad

2211 posts

ZETA_in_olympiad

#48 h Apr 19, 2022, 1:28 PM • 2 Y

Y by ImSh95, Mango247

Edit June 10: added a crucial word.

Claim: $f(0)=0.$
Proof. Let $P(x,y)$ denote the original equation.
$P(0,y)\implies f(f(y))=y+(f(0))^2.$
$P(0,0)\implies f(f(0))=(f(0))^2.$
$P(0,1)\implies f(f(1))=1+(f(0))^2.$
Applying $f$ to both sides of $P(x,y)$ yields,
$x^2+f(y)+(f(0))^2=f(y+(f(x))^2).$
$P(1,f(0))\implies f(f(0)+(f(1))^2)=1+2(f(0))^2.$
$P(f(1),0)\implies f(f(0)+(f(1))^2)=(1+(f(0))^2)^2.$
From this the claim follows. And $f(f(x))=x.$ $\blacksquare$

Claim: $f$ is strictly increasing.
Proof. Let for any reals $a,b$ , $a>b.$
Note that $f(f(\sqrt{a-b})) =\sqrt{a-b} \implies f(\sqrt{a-b}) \neq 0.$
$P(\sqrt{a-b}, b) \implies (f(\sqrt{a-b}))^2+f(b) =f(a).$
And thus our claim follows. $\blacksquare$

Claim: $f(x)=x,$ which in fact holds.
Proof. It's pretty well known and easy to show that the solution of any strictly increasing and involutive function is an identity function. This completes the proof. $\blacksquare$

This post has been edited 2 times. Last edited by ZETA_in_olympiad, Jun 10, 2022, 6:46 AM

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ZETA_in_olympiad

2211 posts

ZETA_in_olympiad

#49 h Jun 10, 2022, 6:41 AM • 1 Y

Y by Mango247

2nd solution

This post has been edited 1 time. Last edited by ZETA_in_olympiad, Jun 10, 2022, 6:56 AM
Reason: e.g.

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th1nq3r

146 posts

th1nq3r

#50 h Jul 21, 2022, 11:01 PM

Y by

Solution

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EurasiaMath

1 post

EurasiaMath

#51 h Nov 28, 2022, 6:19 PM • 1 Y

Y by strong_boy

Video Solution: https://youtu.be/ZZaIPN-ILdk

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SatisfiedMagma

451 posts

SatisfiedMagma

#52 h Apr 26, 2023, 11:33 AM

Y by

Cool problem! I think I understand Cauchy Functional Equations a little bit now.

Solution. The answer is $f \equiv \mathrm{id}$ which clearly works. We will proceed to show that this is unique. Denote
$P(x,y)$ as the assertion to the functional equation.

One can quickly see that $f$ is bijective from $P(0,x)$ . Let $\alpha$ be "the zero" of $f$ . Now we show that $\alpha = 0$ . For this consider the following substitutions.

$P(\alpha, 0)$ gives us $f(\alpha^2 + f(0)) = 0 = f(\alpha)$ . By injectivity, we have that $\alpha^2 + f(0) = \alpha$ .
$P(0, \alpha)$ gives us $f(0) = f(0)^2 + \alpha$ . Now solve the simultaneous equations to get $f(0) = 0$ as desired.

$P(x,0)$ reveals $f(x^2) = f(x)^2 \ge 0$ . So $f$ is positive over positive domain values. $P(0,x)$ gives that $f$ is an involution. Finally observe that $P(x,f(y))$ gives the following partial "Cauchy Condition"
$f(x^2 + y) = f(x^2) + f(y) \implies f(a+b) = f(a) + f(b)$ for all reals $a$ and all $y \ge 0$ . One can extend this partial condition to a full "Cauchy" easily. Observe that for some arbitrary $r \in \mathbb{R}$
$f(0) = f(r-r) = f(r) + f(-r) \iff f(r) = -f(-r)$ which shows $f$ is odd. With the odd nature of $f$ , we can now say that $f$ is completely additive over $\mathbb{R}$ . We will now present two finishes.

Since $f(x) > 0$ for $x > 0$ , $f$ must be strictly increasing. For strictly increasing additive function $f$ , it is well-known that $f(x) = ax$ . Plugging this in would finish the solution.
The second finish uses the involution. Since $f$ is an involution and strictly increasing, $f$ must be the identity function.

So we're done. $\blacksquare$

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ryanbear

1055 posts

ryanbear

#53 h Feb 14, 2024, 5:00 AM

Y by

$f(x^2+f(y))=y+f(x)^2$
Plug in $y=-f(x)^2$ to get $f(x^2+f(-f(x)^2))=0$
Let $u=x^2+f(-f(x)^2)$
$f(u^2+f(y))=y$
so $f(y)=f(f(u^2+f(y)))=u^2+f(y)$
$u^2=0$
$u=0$
$f(0)=0$
Plug in $x=0$ to get $f(f(y))=y+f(0)^2=y$ , so $f$ is injective and surjective
$f(x^2+f(-f(x)^2))=0=f(0)$ , so $x^2+f(-f(x)^2)=0$
$f(-f(x)^2)=-x^2$
$f(f(-f(x)^2))=f(-x^2)=-f(x)^2$
Plug in $y=0$ to get $f(x^2+f(0))=f(x^2)=f(x)^2$ , so $f$ is odd
Plug in $x=1$ to get $f(1)=f(1)^2$ , so $f(1)=0$ or $1$ , but beacuse $f$ is injective, $f(1)=1$
Plug in $y=-y$ to get $f(x^2-f(y))=f(x)^2-y$
$f(x^2+f(y))=y+f(x)^2=f(x)^2+f(f(y))$ , so $f$ is additive
Because $f(x^2)=f(x)^2>0$ , $f$ is bounded over the positives, so $f(x)=kx$
Because $f(x)^2=f(x^2)$ , $k^2x^2=kx^2$ , so $k=0$ or $1$ . Plugging $f(x)=0$ and $f(x)=x$ in, get that only $\boxed{f(x)=x}$ works

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bin_sherlo

663 posts

bin_sherlo

#54 h Nov 22, 2024, 9:31 AM

Y by

$f(x^2+f(y))=y+f(x)^2$ First, by looking at $y$ we see that $f$ is injective, also surjective. We compare $P(x,y)$ and $P(-x,y)$ to get $y+f(x)^2=y+f(-x)^2$ or $|f(x)|=|f(-x)|$ and since $f$ is injective, $f(x)=-f(-x)$ thus, $f$ is odd.
$f(x^2-f(y))+f(x^2+f(y))=-y+f(x)^2+y+f(x)^2=2f(x)^2$ $f$ is surjective hence $f(x^2-y)+f(x^2+y)=2f(x)^2$ and $x=0$ implies $f(0)=0$ . $x=0$ in the original equation gives $f(f(y))=y$ so $f$ is involution. $y=0$ yields $f(x^2)=f(x)^2$ .
$f(x^2+y)=f(y)+f(x)^2=f(y)+f(x^2)$ Let's show that $f(a)+f(b)=f(a+b)$ for all reals. If one of $a,b$ is nonnegative, then $f(x^2)+f(y)=f(x^2+y)$ gives the result. If both of them are negative, then since $f$ is odd $f(-x^2)+f(-y)=f(-x^2-y)$ gives it. Since $f$ is both additive and bounded below on positive values by $f(x^2)=f(x)^2\geq 0$ , this is a Cauchy function with solution $f(x)=cx$ and subsituting back implies $f(x)=x$ as desired. $\blacksquare$

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Saucepan_man02

1299 posts

Saucepan_man02

#55 h Jan 1, 2025, 10:29 AM

Y by

Old-Classic:

Here is the proof:
Let $f(0)=c$ . Then: $P(0, 0): f(c)=c^2$ .
$P(x, f(y))$ : $f(x^2+f(f(y))) = f(y)+f(x)^2 \implies f(x^2+y+c) = f(y)+f(x)^2$ $\implies f(f(x^2+y+c)) = f(f(y)+f(x)^2) \implies x^2+y+2c = y+c^2+(x+c^2)^2$ $\implies 2c = 2c^2 x+c^4+c^2$ Inorder for the above equation to hold for all real $x$ , we have $c=0$ . Thus: $f(f(x))=x, f(x^2)=f(x)^2.$ The latter implies $f(x) \ge 0$ for all $x \ge 0$ . Thus, plugging back $P(x, f(y))$ : $f(x^2+y) = f(x^2)+f(y).$ Note that $f$ is surjective thus, $f(y)$ covers all reals. Thus: $f$ is additive along with $f(x) \ge 0$ for all $x \ge 0$ .

Therefore: $f(x)=ax$ and plugging back $f(x)=x$ is the only solution,

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alexanderhamilton124

378 posts

alexanderhamilton124

#56 h Jan 1, 2025, 11:06 AM • 1 Y

Y by L13832

nice..
solution

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KAME06

116 posts

KAME06

#57 h Wednesday at 9:55 PM

Y by

Saving my solution:
$P(0,0) \Rightarrow f(f(0))=f(0)^2$ . Let $f(0)=k$ , then $f(k)=k^2$ .
(1) $P(0,k) \Rightarrow f(0+f(k))=f(k^2)=k+k^2$ .
(2) $P(k,0) \Rightarrow f(k^2+k)=f(k)^2=k^4$ .
Using (1), $P(k^2, k) \Rightarrow f(k^4+k^2)=k+f(k^2)^2=k+(k+k^2)^2$ .
Using (2), $P(k, k^2+k) \Rightarrow f(k^2+k^4)=k^2+k+k^4$ .
$\Rightarrow k+k^2+2k^3+k^4=k^2+k+k^4 \Rightarrow 3k^3=0 \Rightarrow k=f(0)=0$ .
(3)Now, $P(x, 0) \Rightarrow f(x^2)=f(x)^2$ .
(4) $P(0,x) \Rightarrow f(f(x))=x$ so it's well known that $f$ must be bijective.
(5)Let $c \in \mathbb{R^+}$ . Then $P(\sqrt{c}, 0) \Rightarrow f(c)=f(\sqrt{c})^2 \ge 0$ .
(6)Using (3), (4) and (5): $P(f(\sqrt{c}),-c) \Rightarrow f(f(\sqrt{c})^2+f(-c))=-c+f(f(\sqrt{c}))^2=-c+c=0 \Rightarrow f(\sqrt{c})^2+f(-c)=0 \Rightarrow f(-c)=-f(\sqrt{c})^2=-f(c) \le 0$ .
$P(\sqrt{c}, -c) \Rightarrow f(c-f(-c))=f(c-f(c))=-c+f(c)=-(c-f(c))$ .
If $c-f(c)>0$ , using (5), $-(c-f(c)=f(c-f(c))>0 \Rightarrow c<f(c)$ .
If $c-f(c)<0$ , using (6), $-(c-f(c))=f(c-f(c))<0 \Rightarrow c>f(c)$ . Both contradictions. So $f(c)=c$
Using (6), $f(-c)=-f(c)=-c$ . So $f(x)=x$ .
$f(x^2+f(y))=f(x^2+y)=x^2+y=y+x^2=y+f(x)^2$ .
Solution: $f(x)=x$

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